Question

the 13th term of an ap is four times its 3rd term if its 5th term is 16​

Answer 1

$\bold{Correct\: Question:}$

The 13th term of an AP is four times its 3rd term. If it's 5th term is 16. Find the sum of it's first 10th term.

Solution:

$a_{n}$ = a + (n - 1)d

$a_{13}$ = a + (13 - 1)d

• A.T.Q.

=> $a_{13}$ = 4 × $a_{3}$

=> a + (13 - 1)d = 4[a + (3 - 1)d]

=> a + 12d = 4(a + 2d)

=> a + 12d = 4a + 8d

=> - 3a = - 4d

=> a = $\dfrac{4d}{3}$ ______ (eq 1)

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$a_{5}$ = a + (5 - 1)d

=> 16 = a + 4d

=> 16 = $\dfrac{4d}{3}$ + 4d [From (eq 1)]

=> 16 = $\dfrac{4d\:+\:12d}{3}$

=> 16 × 3 = 16d

=> d = $\dfrac{16\:\times\:3}{16}$

=> d = 3

Put value of d in (eq 1)

=> a = $\dfrac{4(3)}{3}$

=> a = 4

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$S_{n}$ = $\dfrac{n}{2}$ [2a + (n - 1)d]

=> $S_{10}$ = $\dfrac{10}{2}$ [2(4) + (10 - 1)3]

=> 5 [8 + 9(3)]

=> 5 (8 + 27)

=> 5 (35)

=> 175

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Sum of it's first 10th term is 175.

__________ [ANSWER]

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