the sum of the first three terms of an AP is 21 and the sum of the first five terms is 55. find. (1) the first term (2) common difference (3) the sum of the first ten term of the sequence


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Question Number 188758 by Humble last updated on 06/Mar/23

$t h e s u m o f t h e f i r s t t h r e e t e r m s$ $o f a n A P i s 21 a n d t h e s u m o f t h e$ $f i r s t f i v e t e r m s i s 55 . f i n d .$ $(1) t h e f i r s t t e r m$ $(2) c o m m o n d i f f e r e n c e$ $(3) t h e s u m o f t h e f i r s t t e n t e r m$ $o f t h e s e q u e n c e$
	Answered by floor(10²Eta[1]) last updated on 06/Mar/23

	$S_{3} = 21, S_{5} = 55$ $\frac{(a_{1} + a_{3}) 3}{2} = 21, \frac{(a_{1} + a_{5}) 5}{2} = 55$ $\Rightarrow a_{1} + a_{3} = 14, a_{1} + a_{5} = 22$ $\Rightarrow a_{1} + (a_{1} + 2 r) = 14, a_{1} + (a_{1} + 4 r) = 22$ $\Rightarrow a_{1} + r = 7, a_{1} + 2 r = 11$ $\Rightarrow r = 4, a_{1} = 3$ $(1) : 3$ $(2) : 4$ $(3) : S_{10} = \frac{(a_{1} + a_{10}) 10}{2} = 210$
	Answered by Rasheed.Sindhi last updated on 06/Mar/23

	$L e t t i s t h i r d t e r m o f t h e A P$ $(t - 2 d) + (t - d) + t + (t + d) + (t + 2 d) = 55$ $5 t = 55 \Rightarrow t = 11$ $(t - 2 d) + (t - d) + t = 21$ $3 t - 3 d = 21$ $3 (11) - 3 d = 21 \Rightarrow d = \frac{33 - 21}{3} = 4$ $f i r s t t e r m = a = t - 2 d = 11 - 2 (4) = 3$ $S_{10} = \frac{10}{2} [2 (3) + (10 - 1) (4)]$ $= 5 (6 + 36) = 210$
	Answered by Rasheed.Sindhi last updated on 06/Mar/23

	$L e t u & t a r e 2 n d & 3 r d t e r m s r e s p e c t i v e l y .$ $d i s c o m m o n d i f f e r e n c e .$ $∙ (u - d) + u + (u + d) = 21$ $3 u = 21 \Rightarrow u = 7 (2 n d t e r m)$ $∙ (t - 2 d) + (t - d) + t + (t + d) + (t + 2 d) = 55$ $5 t = 55 \Rightarrow t = 11 (3 r d t e r m)$ $∙ d = t - u = 11 - 7 = 4$ $∙ F i r s t t e r m = u - d = 7 - 4 = 3$ $∙ S_{10} = \frac{10}{2} [2 (3) + (10 - 1) (4)]$ $= 5 (6 + 36) = 5 \times 42 = 210$
	Commented by Humble last updated on 07/Mar/23

	$t h a n k s, s i r .$
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