Question

If for an A.P., S15= 147 and s14=123 find t 15

(A) 24 (B) 23 (C) 47 (D) 46
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Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

If for an A.P

$\sf{S_{15} = 147 \: \: and \: \: S_{14} =123 \: \: \: then \: \: t_{15} = } \:$

(A) 24

(B) 23

(C) 47

(D) 46

EVALUATION

Let for the given AP

$\sf{n th \: term = \: t_n \: and \: Sum \: of \: first \: n \: terms = S_n}$

Therefore

$\sf{ S_{15} = t_1 + t_2 + t_3 + .... + t_{14} + t_{15}} \: \: - - (1)$

$\sf{ S_{14} = t_1 + t_2 + t_3 + .... + t_{14} } \: \: - - (2)$

Equation 1 - Equation 2 gives

$\sf{S_{15} - S_{14} = t_{15} } \:$

$\sf{ \implies \: t_{15} = S_{15} - S_{14} } \:$

$\sf{ \implies \: t_{15} = 147 - 123 } \:$

$\sf{ \implies \: t_{15} = 24 } \:$

FINAL ANSWER

Hence the correct option is (A) 24

Answer 2

Question :- if for an A.P., S(15) = 147 and S(14) = 123, find t(15).

(A) 24 (B) 23 (C) 47 (D) 46

Solution :-

given that,

→ sum of first 15 terms of an AP = 147

and,

→ sum of first 14 terms of an AP = 123

so,

→ a1 + a2 + a3 + a4 ____________ a15 = 147 ------- Eqn.(1)

→ a1 + a2 + a3 + a4 ____________ a14 = 123 ------- Eqn.(2)

Subtracting Eqn.(2) from Eqn.(1) , we get,

→ (a1 + a2 + a3 + a4 ____________ a15) - (a1 + a2 + a3 + a4 ____________ a14) = 147 - 123

→ a15 = 24 (A) (Ans.)

Hence, 15th term of given AP will be 24.

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