Question

The sum of 5 th and 9 th terms of an A.P.is 8 and their product is 15. Find the sum of first 28 terms of the A.P.

Answer 1

203 .

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Answer 2

$\large\bold{\underline{\sf{\pink{AnswEr:-}}}}$

$\ \ \ \ \ \ \ \ \bullet$ Sum of 5th and 9th terms of an A.P. is 8.

$\ \ \ \ \ \ \ \ \bullet$ Product of 5th and 9th terms of an A.P. is 15.

$\large\bold{\underline{\sf{\pink{To \; Find :-}}}}$

$\ \ \ \ \ \ \ \ \bullet$ Sum of 28th term of the A.P.

$\large\bold{\underline{\underline{\boxed{\sf{\purple{Solution:-}}}}}}$

★ Sum of 5th and 9th term:-

$\implies \sf a_{5} + \sf a_9 → 8$

• $a + 4d + a + 8d → 8$

• $2a + 12d → 8$

• $2a → 8 - 12d$

• $a → \cancel \dfrac{8 - 12d}{2}$

$\ \ \ \ \ \ \ \implies a → 4 - 6d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ------(1)$

★ Product of 5th and 9th term:-

$\ \ \ \ \ \ \ \implies ( \sf a_5 )( \sf a_9 ) → 15$

$\ \ \ \ \ \ \ \implies (a + 4d)(a + 8d) → 15 \ \ \ \ \ \ \ \ ------(2)$

★ Putting eq.(1) in eq.(2):-

• $(4 - 6d + 4d)(4 - 6d + 8d) → 15$

• $(4 - 2d)(4+2d) → 15$

• $16 - 4{d}^2 → 15$

• $-4{d}^2 → 15 - 16$

• ${d}^2 → \dfrac{1}{4}$

• $\sqrt{{d}^2} → \sqrt{ \dfrac{1}{4}}$

• $\large\bold{\underline{\underline{\boxed{\sf{\red{d → \dfrac{1}{2}}}}}}}$

★ Put value of d in eq.(1):-

• $a → 4 - 6 × \dfrac{1}{2}$

• $a → 4 - 3$

• $\large\bold{\underline{\underline{\boxed{\sf{\red{a → 1}}}}}}$

★ Sum of 28 terms of A.P:-

$\implies \sf S_{28} = \dfrac{28}{2}( 1 + \dfrac{(28 -1)}{2})$

$\ \ \ \ \ \ \ \ \ \ \ \ = 14(1 + \dfrac{27}{2})$

$\ \ \ \ \ \ \ \ \ \ \ \ = 14 × \dfrac{29}{2}$

$\ \ \ \ \ \ \ \ \ \ \ \ = 7 × 29$

$\ \ \ \ \ \ \ \ \ \ \ \ \large\bold{\underline{\underline{\boxed{\sf{\purple{ \sf S_{28} = 203}}}}}}$

$\rule{100}{2}\rule{100}{2}$