FIND THE SUM OF N TERMS FROM THE GIVEN SUM OF SERIES

Question 1 :

If the sum of 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. 

Solution :

Given that :

S₇  =  49

S_n  =  (n/2) [2a + (n - 1)d]

S₇  =  (7/2)[2a + (7-1)d]

(7/2)[2a + 6d]  =  49

2a + 6d  =  49⋅(2/7)

2a + 6d  =  14

Dividing both sides by 2

a + 3d  =  7  ------(1)

S₁₇  =  289

S₁₇  =  (17/2)[2a + (17-1)d]

(17/2)[2a + 16d]  =  289

2a + 16d  =  289⋅(2/17)

2a + 16d  =  34

Dividing both sides by 2

a + 8d  =  17  ------(2)

(1) - (2)

    a + 3d  =  7 

   a + 8d  =  17 

  (-)   (-)   (-)

------------------

-5d  =  -10

d  =  2

By applying the value of d in (1), we get

a + 3(2)  = 7

a + 6  =  7

a  =  7 - 6  =  1

By applying the values of "a" and "d" in the formula for sum of n terms.

S_n  =  (n/2) [2a + (n - 1)d]

S_n  =  (n/2) [2(1) + (n - 1)2]

S_n  =  (n/2) [2 + 2n - 2]

S_n  =  n²

From the General Term Find the Sum of Series

Question 2 :

Show that a₁, a₂,............ an form an AP where an is defined as below

(i) a_n  =  3 + 4n

(ii) a_n  =  9 - 5n

Also find the sum of 15 terms in each case.

Solution :

(i) a_n  =  3 + 4n

From the given general term, we may find the first term and common difference.

By applying n = 1, we get the first term.

First term (a)  =  7, Last term (l)  =  63

Number of terms  =  15

S_n  =  (n/2) [a + l]

S₁₅  =  (15/2) [7 + 63]

=  (15/2) (70)

=  15(35)

  =  525

Hence the sum of 15 terms is 525.

(ii) a_n  =  9 - 5n

From the given general term, we may find the first term and common difference.

By applying n = 1, we get the first term.

First term (a)  =  4,

Last term (l)  =  -66

Number of terms  =  15

S_n  =  (n/2) [a + l]

S₁₅  =  (15/2) [4 - 66]

=  (15/2) [-62]

=  15 (-31)

=  -465

Hence the sum of 15 terms is -465.

Question 3 :

A partial sum of an arithmetic sequence is given. Find the sum 

3 + 7 + 11 + ............. + 39

Solution :

a = 3, d = 7 - 3 ==> 4

Finding the number of terms :

l = 39 = a_n

a + (n - 1)d = 39

3 + (n - 1) (4) = 39

(n - 1)4 = 39 - 3

(n - 1)4 = 36

n - 1 = 9

n = 10

The series consists of 10 terms.

Finding the sum of 10 terms :

Sn = (n/2)[a + l]

S₁₀ = (10/2)[3 + 39]

= 5(42)

= 210

So, the required sum of the series is 210.

Question 4 :

A partial sum of an arithmetic sequence is given. Find the sum 

Solution :

1, -6, -13, ..........

First term = 1

Common difference = -6 - 1 ==> -7

By applying from 0 to 20, the series consists of 21 terms.

To find last term, we apply n = 20

= 1 - 7(20)

= 1 - 140

= -139

S₂₁ = (21/2)[1 - 139]

= (21/2)[- 138]

= 21(-69)

= -1449

So, the required sum is -1449.

Question 5 :

A partial sum of an arithmetic sequence is given. Find the sum 

Solution :

15, 21, 27, ...........

a = 15, d = 21 - 15 ==> 6

The series consists of 6 terms.

By applying the n = 7, we get the last term.

Last term = 6(7) + 3

= 42 + 3

= 45

S₆ = (n/2)[a + l]

= (6/2) [15 + 45]

= 3(60)

= 180

So, the required sum of the series is 180.

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