The 2nd, 4th and 8th term of an AP are the consecutive term of a GP. If the sum of the 3rd and 4th term of the AP is 20. Find the sum of the first four terms of the AP.


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Question Number 61117 by Tawa1 last updated on 29/May/19

$The 2 nd, 4 th and 8 th term of an AP are the consecutive term of a GP .$ $If the sum of the 3 rd and 4 th term of the AP is 20 . Find the sum of the$ $first four terms of the AP .$
	Answered by Kunal12588 last updated on 29/May/19

	$g i v e n a_{2}, a_{4}, a_{8} a r e i n G P$ $\Rightarrow \frac{a_{4}}{a_{2}} = \frac{a_{8}}{a_{4}}$ $\Rightarrow {(a_{4})}^{2} = a_{2} \cdot a_{8}$ $\Rightarrow {(a + 3 d)}^{2} = (a + d) (a + 7 d)$ $\Rightarrow a^{2} + 6 a d + 9 d^{2} = a^{2} + 8 a d + 7 d^{2}$ $\Rightarrow 8 a - 6 a + 7 d - 9 d = 0 \lor d = 0$ $\Rightarrow 2 a - 2 d = 0 \lor d = 0 (1)$ $g i v e n a_{3} + a_{4} = 20$ $a + 2 d + a + 3 d = 20$ $\Rightarrow 2 a + 5 d = 20 (2)$ $f r o m (1) a n d (2)$ $7 d = 20 \Rightarrow d = \frac{20}{7} \lor d = 0$ $a = d = \frac{20}{7} \lor a = 10$ $a_{1} = a = \frac{20}{7} \lor a_{1} = a_{2} = a_{3} = a_{4} = 10$ $a_{2} = a_{1} + d = \frac{40}{7}$ $a_{3} = a_{2} + d = \frac{60}{7}$ $a_{4} = a_{3} + d = \frac{80}{7}$
	Commented by Tawa1 last updated on 29/May/19

	$God bless you sir$
	Answered by Rasheed.Sindhi last updated on 29/May/19

	$AnOther Way$ $a_{2}, a_{4}, a_{8} = a, a r, {a r}^{2} [∵ a_{2}, a_{4}, a_{8} are in GP]$ $a_{4} - a_{2} = a r - a = 2 d \Rightarrow d = \frac{a r - a}{2}$ $a_{8} - a_{4} = {a r}^{2} - a r = 4 d \Rightarrow d = \frac{{a r}^{2} - a r}{4}$ $\frac{a r - a}{2} = \frac{{a r}^{2} - a r}{4}$ $2 r - 2 = r^{2} - r$ $r^{2} - 3 r + 2 = 0 \Rightarrow r = 1, 2$ $∴ r = 2 [r \neq 1 in a GP]$ $∴ d = \frac{a r - a}{2} = \frac{2 a - a}{2} = \frac{a}{2}$ $Now a_{3} + a_{4} = 20 [Given]$ $(a_{2} + d) + a r = 20$ $(a + \frac{a}{2}) + a (2) = 20$ $2 a + a + 4 a = 40$ $a = \frac{40}{7} \Rightarrow d = \frac{a}{2} = \frac{20}{7}$ $a_{1} + a_{2} + a_{3} + a_{4}$ $= (a - d) + a + 20 = 2 a - d + 20$ $= 2 (\frac{40}{7}) - \frac{20}{7} + 20$ $= \frac{80 - 20 + 140}{7} = \frac{200}{7}$
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