a_n is an AP and S_n is sum of n terms of this AP. Given that S_(11) −S_7 =72, determine a_6 +a


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Question Number 175180 by Rasheed.Sindhi last updated on 22/Aug/22

$a_{n} i s a n A P a n d S_{n} i s s u m o f n t e r m s$ $o f t h i s A P .$ $G i v e n t h a t S_{11} - S_{7} = 72, d e t e r m i n e$ $a_{6} + a_{13} .$
	Answered by som(math1967) last updated on 22/Aug/22

	$S_{11} - S_{7} = 72$ $\Rightarrow \frac{11}{2} (2 a_{1} + 10 d) - \frac{7}{2} (2 a_{1} + 6 d) = 72$ $[a_{1} = 1 s t t e r m d = c . d]$ $\Rightarrow 4 a_{1} + 55 d - 21 d = 72$ $\Rightarrow 4 a_{1} + 34 d = 72$ $\Rightarrow 2 a_{1} + 17 d = 36$ $\Rightarrow (a_{1} + 5 d) + (a_{1} + 12 d) = 36$ $\Rightarrow a_{6} + a_{13} = 36$
	Commented by Rasheed.Sindhi last updated on 22/Aug/22

	$T h a n k s s i r!$
	Answered by Rasheed.Sindhi last updated on 22/Aug/22

	$A n O t h e r w a y$ $S_{11} - S_{7} = a_{8} + a_{9} + a_{10} + a_{11}$ $[\overset{\times}{a_{1}} + \overset{\times}{a_{2}} + . . . + \overset{\times}{a_{7}} + a_{8} + a_{9} + a_{10} + a_{11}]$ $= (a + 7 d) + (a + 8 d) + (a + 9 d) + (a + 10 d)$ $= 4 a + 34 d = 2 (2 a + 17 d)$ $= 2 ((a + 5 d) + (a + 12 d))$ $= 2 (a_{6} + a_{13}) = 72$ $a_{6} + a_{13} = 72 / 2 = 36$
	Answered by behi834171 last updated on 22/Aug/22

	$S_{m} - S_{n} = \frac{m}{2} ((m - 1) d + 2 a_{1}) - \frac{n}{2} ((n - 1) d + 2 a_{1}) =$ $= d [\frac{m^{2}}{2} - \frac{n^{2}}{2} - \frac{m}{2} + \frac{n}{2}] + (m - n) a_{1} =$ $= \frac{m - n}{2} [(m + n - 1) d + 2 a_{1}] = \frac{m - n}{m + n} S_{m + n}$ $\Rightarrow S_{m + n} = \frac{m + n}{m - n} (S_{m} - S_{n})$ $\Rightarrow S_{18} = \frac{11 + 7}{11 - 7} \times 72 = 364$ $a_{6} + a_{13} = a_{1} + 5 d + a_{1} + 12 d = 17 d + 2 a_{1} =$ $= \frac{9}{9} [(18 - 1) d + 2 a_{1}] = \frac{S_{18}}{9} = 36 . ◼$
	Commented by Rasheed.Sindhi last updated on 23/Aug/22

	$W o n d e r f u l s i r! T hank you .$
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