# If the Sum of a Certain Number of Terms of the Ap 25, 22, 19, ... is 116. Find the Last Term. - Mathematics

If the sum of a certain number of terms of the AP 25, 22, 19, ... is 116. Find the last term.

#### Solution

The given A.P. is 25,.22,.19.....
Here, a = 25, d = $22 - 25 = - 3$

$S_n = 116$

$\Rightarrow \frac{n}{2}\left[ 2a + \left( n - 1 \right)d \right] = 116$

$\Rightarrow n\left[ 2 \times 25 + \left( n - 1 \right)\left( - 3 \right) \right] = 232$

$\Rightarrow 50n - 3 n^2 + 3n = 232$

$\Rightarrow 3 n^2 - 53n + 232 = 0$

$\Rightarrow 3 n^2 - 29n - 24n + 232 = 0$

$\Rightarrow n(3n - 29) - 8(3n - 29) = 0$

$\Rightarrow (3n - 29)(n - 8) = 0$

$\Rightarrow n = \frac{29}{3} \text { or }8$

$\text { Since n cannot be a fraction, } n = 8 .$

$\text {Thus, the last term }:$

$a_n = a + (n - 1)d$

$\Rightarrow a_8 = 25 + \left( 8 - 1 \right) \times \left( - 3 \right)$

$\Rightarrow a_8 = 4$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 19 Arithmetic Progression
Exercise 19.4 | Q 26 | Page 31