Question

answer property please don't Spam​

Answer 1

⟶ Questíon :-

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

⟶ Gíven :-

• A.P = 25, 22, 19, ........
• First term, a = 25
• Common difference, d= 22 - 25 = -3

⟶ To Fínd :-

The last term of the given A.P.

⟶ Solutíon :-

$\sf sum \: of \: n \: terms = \boxed{ \bf S_{n} = \frac{n}{2}(2a + (n - 1)d }$

$: \Longrightarrow \sf \frac{n}{2} (50 + ( n- 1)( - 3) = 116 \\ \\ : \Longrightarrow \sf n(50 - 3n + 3) = 232 \\ \\ : \Longrightarrow \sf - 3 {n}^{2} + 53n - 232 = 0 \\ \\ : \Longrightarrow \sf {3n}^{2} - 53n + 232 = 0 \\ \\ : \Longrightarrow \sf {3n}^{2} - 29n - 24n + 232 = 0 \\ \\ : \Longrightarrow \sf n(3n - 29) - 8(3n - 29) = 0 \\ \\ : \Longrightarrow \sf (n - 8)(3n - 29) = 0 \\ \\ : \Longrightarrow \sf n = 8(or)n = \frac{29}{3}$

$\therefore$n = 8

Last term = $\sf T_{8}$= a + (n - 1)d

=$\sf T_{8}$= 25 + (7)(-3) = 25 - 22 = 4

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