One of the most important and exam oriented question from Chapter name- Arithmetic Progression

Class 10th

Chapter number- 9

Exercise :- 9.6

This type of question has been asked in previous years exams.

In this question we have been given that the sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161

Now we have to find the 28th term of this A.P.

CBSE DHANPAT RAI publication

CBSE Mathematics Class 10th

Question 30

This is the video solution for this question. Thank you!!

Sum of first 7 terms of an A.P., S

_{7}= 63.And sum of next 7 terms is 161.

So, the sum of first 14 terms, S

_{14}= Sum of first 7 terms + Sum of next 7 termsS

_{14}= 63 + 161 = 224By using the formula of the sum of n terms of an A.P.

S

_{n}= n[2a + (n − 1)d] / 2.So, S

_{7}= 7(2a + (7 − 1)d) / 2=> 7(2a + 6d) / 2 = 63

=> 2a + 6d = 18 . . . . (1)

Also, S

_{14}= 14(2a + (14 − 1)d) / 2=> 14(2a+13d)/2 = 224

=> 2a+13d = 32 . . . . (2)

Now, subtracting eq(1) from eq(2), we get

=> 13d – 6d = 32 – 18

=> 7d = 14

=> d = 2

On putting d = 2 in eq(1), we get,

=> 2a + 6(2) = 18

=> 2a = 18 – 12

=> a = 3

Thus, a

_{28}= a + (28 – 1)d = 3 + 27 (2) = 3 + 54 = 57Hence, the 28th term is 57.