Question

HIMETIC PROGRESSIONS
How many terms of the AP:9,17,25,... must be taken to give a sum of 636?
plss anwer my question very fast​

Answer 1

Answer:

hope this is helpful.

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

• AP: 9,17,25
• S_n = 636

${\bf{\blue{\underline{To\:Find:}}}}$

• n = ?

${\bf{\blue{\underline{Formula\:Used:}}}}$

$: \dagger \: \boxed {\sf{ s_{n} = \frac{n}{2} [2a + (n - 1)d]}}$

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{ s_{n} = \frac{n}{2} [2 \times 9 + (n - 1)8]}}$

$: \implies{\sf{ 636 = \frac{n}{2} [18+ 8n - 8]}}$

$: \implies{\sf{ 1272 = n(10 + 8n)}}$

$: \implies{\sf{ 1272 = 10n + 8 {n}^{2} }}$

$: \implies{\sf{ 8 {n}^{2} + 10n - 1272 = 0 }}$

$: \implies{\sf{ 8 {n}^{2} + 106n - 96n - 1272 = 0 }}$

$: \implies{\sf{ 2n(4n + 53) - 24(4n + 53) = 0}}$

$: \implies{\sf{ (4n + 53)(2n - 24) = 0}}$

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$\longmapsto{\sf{ (4n + 53) = 0}}$

$\longmapsto{\sf{ 4n =- 53 }}$

$\longmapsto{\sf{ n = \frac{-53}{4} }}$

n can never be negative so, we have to cancel this value.

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$\longmapsto{\sf{ 2n - 24 = 0}}$

$\longmapsto{\sf{ 2n = 24 }}$

$\longmapsto{\sf{ n = \frac{24}{2} }}$

$\longmapsto{\sf{ n = </strong><strong>1</strong><strong>2</strong><strong> }}$

• Hence n = 12