Basic Mathematics
1.
Arithmetic Progression
Geometric Progression
An arithmentic Progression is a sequence in which the difference between two consecutive forms like the nth and (n+1) th term is constant.
A geometric progression in which the ratio of each term to the one before it is constant. The ratio is called the common ratio.
i) 3, 6, 9, 12, 15, .....................................
ii) 1,
..............................
i) 3, 6. 12, 24, 48, ............................... ii) 2, 1,
Formula ; (i) a+(n-1)d
ii)
Formula ; (i) nth term Tn = ar n-1
(ii) Sn =
2. Given r th term of arithmetical Progression is n and nth term is r.
Soln : Let, the first term is a and general difference is d.
\ rth term = a + (r-1) d
\ nth term = a + (n-1)d
According to the question,
a + (r-1) d= n
= a + rd – d = n ......................... (ii)
and a + (n-1) d = r
= a + nd – d = r ------------------- (ii)
Now from (ii) - (i) we get,
a + nd – d – (a + rd – d) = r – n
= a + nd – d – a – rd + d = r- n
= nd – rd = r – n
= d ( n- r) = r – n
= d (n – r ) = r – n
= d ( n – r) = - (n-r)
= d =
\ d= -1
Now from (i), we got
a + rd – d = n
=> a + r (1) – (-1) = n
=> a – r + 1 = n
\ a = n + r -1
So, the mth term is = at (m-1) d
= n + r – 1+m-1 (-1)
= n + r – 1 – m + 1
= r + n – m
(Showed)
3. The sum of three number is a G..P. is 38 and their product is 1728. Find the numbers.
Soln
Let the number be
Their product
=> a3 = 1728
=> a = 12
So the numbers are
The sum of these numbers = 38
=>
=>
=> 12r2 + 12r + 12 = 38r
=> 12r2 + 12r – 38r + 12 = 0
=> 12r – 26r + 12 = 0
=> 2 (6r2 - 13r + 6) = 0
=> 6r2 -13r + 6 = 0
=> 6r2 – 9r - 4r + 6 = 0
=> 3r (2r – 3 ) -2 ( 2r-3) = 0
=> (3r-2) (2r-3) = 0
\ 3r- 2 = 0 2 r – 3 = 0
\ r =
\ r = 
r = gives the numbers us =
= 12 ´
= 18, 12, 8
r = gives the number as =
,
= 12 ´
12, 12,
= 8, 12, 18
Which same as the first set
Hence the required rumbers are 8, 12, 18 Ans
4. Find the sum of first 10 terms of a G. P. given by 1,
Soln : Here, the first term is a = 1
Common ratio, r = -
= - < 1
and n = 10 (given)
\The sum of first 10 terms
S10 =
=
=
=
=
=
=
Ans :
5. The rate of monthly salary of a person increased annually in A. P. It is known that he was darning TK. 200 and TK. 380 a month during 11th and 29th year of his service respectively. What was his starting alary and what was his total earnings during 29 years of his service ?
Soln- Let,
T11 = 200 = a + (11-1) d ................... (i)
T 29 = 380 = a + (29-1) d ................. (ii)
Starting salary a = ? Total earnings during 29 years T 29 ´ 12 = ?
\ S29 =
{2a + (n-1)d}
=
{2a + (29-1)d } .................................. (iii)
ii)- (i) 180 = 28d- 10d
= 180 = 18d
= 18d = 180
= d =
\ d= 10
d = 10, we have get (i)
200 = a + (11-1) 10
=> 200 = a + 100
=> a + 100 = 200
=> a = 200-100
\ a = 100
Starting salary is 100 taka
S29 ={2´100 + (29-1)10}
= (200 + 280)
=´480
= 6960
\ Total earnings during 29 years S29 = S29 ´ z
= 6960 ´ 12
= 83,520 taka
Ans : 83,520 taka
6. A class consists of a number of boys whose ages are in arithmetical progression, the common difference being 6 months. If the youngest boy of the class be only 10 years old and sum of the ages of all the boy’s in the class be 169 years. Find the number of boys and the age of the oldest boy in the class.
Sol : The first term of the A. P. is 10, i. e, a1 and the common difference is
or 6 months.
Let, n be the required number of boys. using the formula sn=
{2a+(n-1)d }
In this case sn = 169, a= 10, d =
Substituting the given values, we have
169 ={2.10 + (n-1) }
=> 169 =(20 + 
)
=> 169 =
=> 169 = (
=> 169 =
=> n2 + 39n = 676
=> n2 + 39n - 676 = 0
=> n2 + 52n - 13n - 676 = 0
=> n (n+52)-13 (n+52)=0
=> (n+52) (n-13) = 0
Hence n=13, or, n=-52, the negative value of n being inadmissible.
Therefore the number of boys in the class in 13.
The age of oldest boy in the class i. e. the 13th term of the series is,
an = a + (n-1) d
= 10+(13-1)
= 10 +12 ´ = 16 years Ans
7. Find the sum of the first 17 terms of the geometric progression
9, -3, 1, -
.................
Sol : Using the formula sn =
In this case, a = 9, r = -, n = 17
Substituting the given value, we have
Sn =
=
=
=
=
(APP) Ans
8. Sum to n terms the series 7 + 77 + 777 + ..................
Solv : Sn = 7 ( 1+11+111 + ...................... to n terms)
=
[ 9 + 99 + 999 + ...................... to n terms ]
=[(10-1) + (102-1) + (103-1) + .............. to n terms ]
= [ (10 + 102 + 103 + ........................ to n terms)- (1+11+111+ ........ to n
terms)]
=[(10 + 102 + 103 + ....................... + 10n)-n]
=
=
=
Ans
9. A man secures an interest free loan of TK 14,500 from a friend and agrees to repay it in ten instilments. He pays TK. 1,000 as first instilment and then increases each instatement by equal amount over the preceding instilment. What will be his last instilment ?
Solv : We know, Sn =
{2a+(n-1)d}
=> 14500 =
{2.1000 + (10-1) d }
=> 14500 = 5 (2000 + 9d)
=> 14500 = 10,000 + 45d
=> 45d = 4500
\ d = 100
So the last instilment is, an = a+(n-1) d
a10 = 1000 + (10-1)100
= 1000 + 900
= TK 1900 Ans
1.
Arithmetic Progression
Geometric Progression
An arithmentic Progression is a sequence in which the difference between two consecutive forms like the nth and (n+1) th term is constant.
A geometric progression in which the ratio of each term to the one before it is constant. The ratio is called the common ratio.
i) 3, 6, 9, 12, 15, .....................................
ii) 1,
.............................. i) 3, 6. 12, 24, 48, ............................... ii) 2, 1,
Formula ; (i) a+(n-1)d
ii)
Formula ; (i) nth term Tn = ar n-1
(ii) Sn =
2. Given r th term of arithmetical Progression is n and nth term is r.
Soln : Let, the first term is a and general difference is d.
\ rth term = a + (r-1) d
\ nth term = a + (n-1)d
According to the question,
a + (r-1) d= n
= a + rd – d = n ......................... (ii)
and a + (n-1) d = r
= a + nd – d = r ------------------- (ii)
Now from (ii) - (i) we get,
a + nd – d – (a + rd – d) = r – n
= a + nd – d – a – rd + d = r- n
= nd – rd = r – n
= d ( n- r) = r – n
= d (n – r ) = r – n
= d ( n – r) = - (n-r)
= d =
\ d= -1
Now from (i), we got
a + rd – d = n
=> a + r (1) – (-1) = n
=> a – r + 1 = n
\ a = n + r -1
So, the mth term is = at (m-1) d
= n + r – 1+m-1 (-1)
= n + r – 1 – m + 1
= r + n – m
(Showed)
3. The sum of three number is a G..P. is 38 and their product is 1728. Find the numbers.
Soln
Let the number be
Their product
=> a3 = 1728
=> a = 12
So the numbers are
The sum of these numbers = 38
=>
=>
=> 12r2 + 12r + 12 = 38r
=> 12r2 + 12r – 38r + 12 = 0
=> 12r – 26r + 12 = 0
=> 2 (6r2 - 13r + 6) = 0
=> 6r2 -13r + 6 = 0
=> 6r2 – 9r - 4r + 6 = 0
=> 3r (2r – 3 ) -2 ( 2r-3) = 0
=> (3r-2) (2r-3) = 0
\ 3r- 2 = 0 2 r – 3 = 0
\ r =
\ r = r = gives the numbers us =
= 12 ´
= 18, 12, 8
r = gives the number as =
, = 12 ´
12, 12, = 8, 12, 18
Which same as the first set
Hence the required rumbers are 8, 12, 18 Ans
4. Find the sum of first 10 terms of a G. P. given by 1,
Soln : Here, the first term is a = 1
Common ratio, r = -
= - < 1 and n = 10 (given)
\The sum of first 10 terms
S10 =
=
=
=
=
=
=
Ans :
5. The rate of monthly salary of a person increased annually in A. P. It is known that he was darning TK. 200 and TK. 380 a month during 11th and 29th year of his service respectively. What was his starting alary and what was his total earnings during 29 years of his service ?
Soln- Let,
T11 = 200 = a + (11-1) d ................... (i)
T 29 = 380 = a + (29-1) d ................. (ii)
Starting salary a = ? Total earnings during 29 years T 29 ´ 12 = ?
\ S29 =
{2a + (n-1)d} =
{2a + (29-1)d } .................................. (iii) ii)- (i) 180 = 28d- 10d
= 180 = 18d
= 18d = 180
= d =
\ d= 10
d = 10, we have get (i)
200 = a + (11-1) 10
=> 200 = a + 100
=> a + 100 = 200
=> a = 200-100
\ a = 100
Starting salary is 100 taka
S29 =
{2´100 + (29-1)10} =
(200 + 280) =
´480 = 6960
\ Total earnings during 29 years S29 = S29 ´ z
= 6960 ´ 12
= 83,520 taka
Ans : 83,520 taka
6. A class consists of a number of boys whose ages are in arithmetical progression, the common difference being 6 months. If the youngest boy of the class be only 10 years old and sum of the ages of all the boy’s in the class be 169 years. Find the number of boys and the age of the oldest boy in the class.
Sol : The first term of the A. P. is 10, i. e, a1 and the common difference is
or 6 months. Let, n be the required number of boys. using the formula sn=
{2a+(n-1)d } In this case sn = 169, a= 10, d =
Substituting the given values, we have
169 =
{2.10 + (n-1) } => 169 =
(20 + ) => 169 =
=> 169 =
(=> 169 =
=> n2 + 39n = 676
=> n2 + 39n - 676 = 0
=> n2 + 52n - 13n - 676 = 0
=> n (n+52)-13 (n+52)=0
=> (n+52) (n-13) = 0
Hence n=13, or, n=-52, the negative value of n being inadmissible.
Therefore the number of boys in the class in 13.
The age of oldest boy in the class i. e. the 13th term of the series is,
an = a + (n-1) d
= 10+(13-1)
= 10 +12 ´
= 16 years Ans 7. Find the sum of the first 17 terms of the geometric progression
9, -3, 1, -
................. Sol : Using the formula sn =
In this case, a = 9, r = -
, n = 17 Substituting the given value, we have
Sn =
=
=
=
=
(APP) Ans 8. Sum to n terms the series 7 + 77 + 777 + ..................
Solv : Sn = 7 ( 1+11+111 + ...................... to n terms)
=
[ 9 + 99 + 999 + ...................... to n terms ] =
[(10-1) + (102-1) + (103-1) + .............. to n terms ] =
[ (10 + 102 + 103 + ........................ to n terms)- (1+11+111+ ........ to n terms)]
=
[(10 + 102 + 103 + ....................... + 10n)-n] =
=
=
Ans 9. A man secures an interest free loan of TK 14,500 from a friend and agrees to repay it in ten instilments. He pays TK. 1,000 as first instilment and then increases each instatement by equal amount over the preceding instilment. What will be his last instilment ?
Solv : We know, Sn =
{2a+(n-1)d} => 14500 =
{2.1000 + (10-1) d } => 14500 = 5 (2000 + 9d)
=> 14500 = 10,000 + 45d
=> 45d = 4500
\ d = 100
So the last instilment is, an = a+(n-1) d
a10 = 1000 + (10-1)100
= 1000 + 900
= TK 1900 Ans
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