Let $$a1$$,$$a2$$,…,$$a30$$ be an A.P., $$S=$$∑$$i=130$$ ai and $$T=$$∑$$i=115$$ $$a(2i$$−$$1)$$. If $$a5=27$$ and S−$$2T=75$$ then $$a10$$ is

$$S=a1+a2+$$...$$+a30=152a1+29dT=a1+a3+$$…$$+a29T=1522a1+14$$×$$2d$$−−−$$12T=152a1+28d$$−−−$$2S$$−$$2T=15d=75d=5a10=a5+5d=27+25=52$$

Arithmetic progression

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Question

$Let a_{1}, a_{2}, \dots, a_{30} be an A.P., S = \sum_{i = 1}^{30} a_{i} and T = \sum_{i = 1}^{15} a_{(2 i - 1)} . If a_{5} = 27 and S - 2 T = 75 {then a}_{10} is$

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Solution

${S=a}_{1} + a_{2} + ... {+a}_{30}$
$\begin{array}{l} = 15 (2 a_{1} + 29 d) \\ T = a_{1} + a_{3} + \dots + a_{29} \\ T = \frac{15}{2} [2 a_{1} + 14 \times 2 d] - - - (1) \\ 2 T = 15 [2 a_{1} + 28 d] - - - (2) \\ S - 2 T = 15 d = 75 \\ d = 5 \\ a_{10} = a_{5} + 5 d \\ = 27 + 25 = 52 \end{array}$

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