Let T_k is the k^(th) term and S_k is the sum of the first k term in arithmetic progression If T_3 + T_6 + T_9 + T_(12) + T_(15) + T_(18) = 45 Find S


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Question Number 24677 by Joel578 last updated on 24/Nov/17

$Let T_{k} is the k^{th} term and S_{k} is the sum$ $of the first k term in arithmetic progression$ $If T_{3} + T_{6} + T_{9} + T_{12} + T_{15} + T_{18} = 45$ $Find S_{20}$
	Answered by jota+ last updated on 24/Nov/17

	$6 T_{1} + (2 + 5 + 8 + 11 + 14 + 17) r = 45$ $2 T_{1} + 19 r = 15$ $T_{1} + (T_{1} + 19 r) = 15$ $T_{1} + T_{20} = 15$ $S_{20} = 15 \times 10$
	Commented by Joel578 last updated on 22/Dec/17

	$t h a n k y o u v e r y m u c h$
	Answered by ajfour last updated on 22/Dec/17

	$3 (T_{3} + T_{6} + T_{9} + . . . + T_{18}) = 135$ $\Rightarrow (T_{2} + T_{3} + T_{4}) + (T_{5} + T_{6} + T_{7}) + . .$ $. . . . + (T_{17} + T_{18} + T_{19}) = 135 . . (i)$ $a n d (\frac{T_{1} + T_{20}}{2}) \times 20 = S_{20}$ $\Rightarrow T_{1} + T_{20} = \frac{S_{20}}{10} . . . . (i i)$ $a d d i n g (i) a n d (i i)$ $S_{20} = 135 + \frac{S_{20}}{10}$ $\Rightarrow S_{20} = 135 \times \frac{10}{9} = 150 .$
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