Question

sum of 4 consecutive terms which are in AP is 32 and the ratio of the first and last terms to the product of two middle terms is 7/15 find the number ​

Answer 1

Solution:

Given:

• Sum of four consecutive numbers of an A.P. is 32.

• The ratio of the first and last terms to the product of two middle terms is 7:15.

To Find:

• Find the numbers.

Let first term be 'a' and common difference be 'd'.

And four consecutive terms of A.P. be (a), (a + d), (a + 2d) and (a + 3d).

$\longrightarrow \sf a+(a+d)+(a+2d)+(a+3d)=32\\ \\ \\ \longrightarrow \sf 4a +6d=32\\ \\ {\underline{\bf Now,\; take\;2\;common\;from\;above\;equation.}}\\ \\ \longrightarrow \sf 2a+3d=16\\ \\ \\ \longrightarrow a=\dfrac{16-3d}{2}\;\;\;\;\;\;..........(1)\\ \\ \\ {\underline{\bf And,\;ratio\;of\;first\;and\;last\;terms\;to\;the\;product\;of\;two\;middle\;terms\;is\;7:15.}}\\ \\ \\ \longrightarrow \sf \dfrac{a(a+3d)}{(a+d)(a+2d)}=\dfrac{7}{15}\\ \\ \\ \longrightarrow \sf 15a(a+3d)=7(a+d)(a+2d)$

$\longrightarrow \sf 15a^{2}+45ad=7a^{2}+21ad+14d^{2}\\ \\ \\ \longrightarrow \sf 8a^{2}+24ad-14d^{2}=0\\ \\ {\underline{\bf Now,\;again\;take\;2\;common\;from\;above\;equation,}}\\ \\ \\ \longrightarrow \sf 4a^{2}+12ad-7d^{2}=0\\ \\ \\ \longrightarrow \sf 4a^{2}+14ad-2ad-7d^{2}=0\\ \\ \\ \longrightarrow \sf 2a(2a+7d)-d(2a+7d)=0\\ \\ \\ \longrightarrow \sf (2a-d)(2a+7d)=0\\ \\ \\ \longrightarrow \sf a=-\dfrac{7d}{2}\;and\;a=\dfrac{d}{2}\;\;\;\;\;\;..........(2)$

${\underline{\bf Substitute\;values\;of\;a\;from\;equation\;(1)\;in\;equation\;(2)\;,we\;get:}} \\ \\ \\  \bf Case\;I:When\;a=-\dfrac{7d}{2},\\ \\ \\ \longrightarrow \sf a=\dfrac{16-3d}{2}\\ \\ \\ \longrightarrow \sf -\dfrac{7d}{2}=\dfrac{16-3d}{2}\\ \\ \\ \longrightarrow \sf -7d=16-3d\\ \\ \\ \longrightarrow \sf -4d=16\\ \\ \\ \longrightarrow \sf d=-4\\ \\ \\ \sf So,\; a=-\dfrac{7\times (-4)}{2}\\ \\ \\ \longrightarrow \sf a=14$

${\underline{\bf Thus,\;four\;numbers\;are:}} \\ \\ \\ \longrightarrow \sf a_{1}=14\\ \\ \\ \longrightarrow \sf a_{2}=14-4=10\\ \\ \\ \longrightarrow \sf a_{3}=14-8=6\\ \\ \\ \longrightarrow \sf a_{4}=14-12=2.$

${\bf{Case\;II: When\;a=\dfrac{d}{2}}} \\ \\ \\ \longrightarrow \sf a=\dfrac{16-3d}{2}\\ \\ \\ \longrightarrow \sf \dfrac{d}{2}=\dfrac{16-3d}{2}\\ \\ \\ \longrightarrow \sf d=16-3d\\ \\ \\ \longrightarrow \sf 4d=16\\ \\ \\ \longrightarrow \sf d = 4\\ \\ \\ So,\;a=\dfrac{4}{2}=2$

${\underline{\bf Thus,\;four\;numbers\;are:}}\\ \\ \\ \longrightarrow \sf a_{1}=2\\ \\ \\ \longrightarrow \sf a_{2}=2+4=6\\ \\ \\ \longrightarrow \sf a_{3}=2+8=10\\ \\ \\ \longrightarrow \sf a_{4}=2+12=14.$

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