Question

If 8 times the 8th term of an AP is equal to 15 times its 15th term then find the 23rd term.

Answer 1

$\textsf{Concept:}$

$\textsf{The n th term of the A.P a, a+d, a+2d,........ is}$

$\boxed{\mathsf{t_n=a+(n-1)d}}$

$\textsf{Given:}$

$\mathsf{8\;t_8=15\;t_{15}}$

$\implies\mathsf{8(a+7d)=15(a+14d)}$

$\implies\mathsf{8a+56d=15a+210d}$

$\implies\mathsf{7a+154d=0}$

$\implies\mathsf{a+22d=0}$

$\implies\mathsf{a+(23-1)d=0}$

$\implies\boxed{\mathsf{t_{23}=0}}$

Find more:

In the AP 7,14,21... how many terms are to be considered for getting sum 5740.

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Answer 2

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GIVEN:

In an AP ,,,

8 t8 = 15 t15

TO FIND :

The 23th term of an AP, t23 = ?

SOLUTION:

The general term,tn

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

where a is the first term

d is the common difference

n is the nth term of an AP.

8 t8 = 15 t15

8 [ a + ( 8 - 1) d ] = 15 [ a + ( 15 - 1 ) d ]

8 [ a + 7d ] = 15 [ a + 14d]

8a + 56d = 15a + 210 d

8a - 15a = 210 d - 56d

-7a = 154 d

a / d = -154 / 7

a / d = - 22

d = -a /22

SUBSTITUTE THE VALUE OF d ,,

t23 = a + ( 23 -1) d

= a + 22d

= a + 22 ( - a /22)

= a - a

t23 = 0

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