# FINDING THE SUM OF GIVEN NUMBER OF TERMS FROM THE GIVEN INFORMATION

Finding the Sum of Given Number of Terms from the Given Information :

In this section, we will see some example problems with mixed information based on the topic arithmetic series.

To find the sum of an arithmetic series, we use the formulas give below.

Sn  =  (n/2) [a + l] (or)

Sn  =  (n/2) [2a + (n - 1)d]

nth term of arithmetic progression :

an  =  a + (n - 1)d

Total number of terms :

n  =  [(l - a)/d] + 1

a = first term, d = common difference, n = number of terms and l = last term.

## Finding the Sum of Given Number of Terms from the Given Information - Examples

Question 1 :

Given a = 7, a13 = 35 find d and S13

Solution :

13th term of the arithmetic progression,

a13  =  35

a + 12d  =  35 ---(1)

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

7 + 12d  =  35

12d  =  35 - 7

12d  =  28

d  =  28/12  =  14/6  =  7/3

S =  (n/2) [2a + (n - 1)d]

S13  =  (13/2) [2(7) + (13 - 1)(7/3)]

S13  =  (13/2) [14 + 12(7/3)]

S13  =  (13/2) [14 + 28]

S13  =  (13/2) (42)

S13  =  273

Hence, the sum of 13 terms of the arithmetic progression is 273.

Question 2 :

Given a12 = 37, d = 3 find a and S12

Solution :

12th term of AP :

a12 = 37

a + 11d  =  37 -----(1)

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

a + 11(3)  =  37

a + 33  =  37

a  =  37 - 33  =  4

Now, we find the sum of 12 terms

sn  =  (n/2)[2a + (n - 1|)d]

s12  =  (12/2)[2(4) + (12 - 1)(3)]

s12  =  6[8 + 11(3)]

s12  =  246

Hence, the sum of first 12 terms is 246.

Question 3 :

Given a3 = 15 , S10 = 125 find d and a10

Solution :

Here, we have 3rd term and sum of first 10 terms. By using these information, we should find the common difference and 10th term.

a3 = 15

a + 2d  =  15  ----(1)

S10 = 125

Sn  =  (n/2)[2a + (n - 1)d]

Here, n = 10

(10/2)[2a + (10 - 1)d]  =  125

5[2a + 9d]  =  125

Dividing each side by 5.

2a + 9d  =  25  ----(2)

By solving (1) and (2), we will get the values of a and d.

(1) ⋅ 2 - (2)

2a + 4d  =  10

2a + 9 d  =  25

(-)     (-)     (-)

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

-5d  =  -15

d  =  3

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

a + 2(3)  =  15

a + 6  =  15

a  =  15 - 6  =  9

10th term :

a10  =  a + 9d

a10  =  9 + 9(3)

=  9 + 27

a10  =  36

Hence, the 10th term is 36. After having gone through the stuff given above, we hope that the students would have understood, finding the sum of given number of terms from the given information

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