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# Question 9. If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, . . ., is 116. Find the last term.

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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 an arithmetic progression such that sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, . . ., is 116.

Now we have to find the last term.

CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 9

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1. Given A.P. has first term(a) = 25,

Common difference(d) = 22 â€“ 25 = -3 and sum(Sn) = 116.

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

=> 116 = n[2(25) + (n âˆ’ 1)(âˆ’3)]/2

=> n[53 âˆ’ 3n]/2 = 116

=> 53n â€“ 3n2Â = 232

=> 3n2Â â€“ 53n + 232 = 0

=> 3n2Â â€“ 24n â€“ 29n + 232 = 0

=> 3n(n â€“ 8) â€“ 29 (n â€“ 8) = 0

=> (3n â€“ 29)( n â€“ 8 ) = 0

=> n = 29/3 or n = 8

Ignoring n = 29/3 as number of terms cannot be a fraction, so we get, n = 8.

So, the last term is:

a8Â = a + (8 â€“ 1)d = 25 + 7(-3) = 25 â€“ 21 = 4

Hence, the last term of the given A.P. is 4.

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