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# Question 28. The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1: 2. Find the first and 15th term of the A.P.

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This is an arithmetic progression based question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise – 9.6
In this question we have been given that the sum of first 9 terms of an A.P. is 162.

Also The ratio of its 6th term to its 13th term is 1: 2.

Now we have to Find the first and 15th term of the A.P

CBSE DHANPAT RAI PUBLICATIONS
Understanding CBSE Mathematics
Class :- 10th
Question no 28

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1. We know sum of n terms of an A.P. is given by SnÂ = n[2a + (n âˆ’ 1)d] / 2.

Therefore, Sum of first 9 terms of given A.P. = S9Â = 9[2a + (9 âˆ’ 1)d] / 2 = 162

=> 162 = 9(2a + (9 âˆ’ 1)d) / 2

=> 2a + 8d = 36

=> a + 4d = 18 Â â€¦..(1)

By using the formula of nth term of an A.P.

anÂ = a + (n â€“ 1)d

So,

Given a6Â : a13Â = 1 : 2,

=> a13Â = 2a6

=> a+12d = 2(a + 5d)

=> a+12d = 2a + 10d

=> a = 2d â€¦..(2)

On putting (2) in (1), we get,

=> 2d + 4d = 18

=> 6d = 18

=> d = 3

On putting d = 3 in (2), we get,

a = 2(3) = 6, which is the first term.

Now 15th term, a15Â = a + 14d = 6 + 14 Ã— 3 = 6 + 42 = 48

Hence, the first and 15th term of the A.P. are 6 and 48 respectively.

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