Let a1- a2- - a30 be an A-P- S -i-130 ai and T - -i-115 a2i - 1- If a5 - 27 and S - 2T - 75- then a10 is equal to -

Given : nbsp; a_1, a_2, ......, a_30 is an A.P. S =∑_i=1^30 a_i nbsp; T = ∑_i=1^15 a_2i 1 nbsp; a_5 = 27 and S 2T = 75 Now assuming the first term as ...

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Byju's Answer

Standard XII

Mathematics

Common Difference

Let a1, a2, ....

Question

Let $a_{1}, a_{2}, . . . . . ., a_{30}$ be an A.P., $S = 30 \sum i = 1 a_{i}$ and $T = 15 \sum i = 1 a_{2 i - 1}$ . If $a_{5} = 27$ and $S - 2 T = 75$ , then $a_{10}$ is equal to :

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Solution

The correct option is D 52
Given :
$a_{1}, a_{2}, . . . . . ., a_{30}$ is an A.P.
$S = 30 \sum i = 1 a_{i}$
$T = 15 \sum i = 1 a_{2 i - 1}$
$a_{5} = 27$ and $S - 2 T = 75$
Now assuming the first term as 'a' and the common difference to be 'd',
$S = a_{1} + a_{2} + a_{3} + . . . . . . . + a_{30}$
$T = a_{1} + a_{3} + a_{5} + . . . . . . . + a_{29}$

Writing $S - 2 T$ and rearranging the terms, we will get
$S - 2 T = (a_{2} - a_{1}) + (a_{4} - a_{3}) + (a_{6} - a_{5}) + . . . . . . . (a_{30} - a_{29}) = 75$
$\Rightarrow 15 \times d = 75 \Rightarrow d = 5$
As given,
$a_{5} = 27 \Rightarrow a + 4. d = 27$
$\Rightarrow a = 7$
$∴ a_{10} = a + 9 d = 7 + 9 \times 5 = 52$

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Let a1,a2,......,a30 be an A.P., S=30∑i=1ai and T=15∑i=1a2i−1. If a5=27 and S−2T=75, then a10 is equal to :

Let $a_{1}, a_{2}, . . . . . ., a_{30}$ be an A.P., $S = 30 \sum i = 1 a_{i}$ and $T = 15 \sum i = 1 a_{2 i - 1}$ . If $a_{5} = 27$ and $S - 2 T = 75$ , then $a_{10}$ is equal to :