Question

The sum of three consecutive terms of an A.P. is 15 and sum of their squares is 83 . Find the terms

Answer 1

I hope it is helpful for you

Answer 2

Answer:

$Required \: 3\: consecutive \\terms \: in \: given \:A.P: \\ 3,5,7 \: Or \: 7,5,3$

Step-by-step explanation:

$Let \: (a-d),a\: and \:(a+d) \\are \: three \: consecutive\: terms\\in \: A.P$

According to the problem given,

$i) Sum \: of \: 3 \: terms = 15$

$\implies (a-d)+a+(a+d)=15$

$\implies a-d+a+a+d = 15$

$\implies 3a = 15$

$\implies a =\frac{15}{3}$

$\implies a = 5---(1) [/tex ]</p><p>[tex]ii) Sum \: of \: squares = 83$

$\implies (a-d)^{2}+a^{2}+(a+d)^{2}=83$

$\implies (a-d)^{2}+(a+d)^{2}+a^{2}=83$

$\implies 2(a^{2}+d^{2})+a^{2}=83$

/*By algebraic identities :

(x+y)²+(x-y)² = 2(x²+y²) */

$\implies 2a^{2}+2d^{2}+a^{2}=83$

$\implies 3a^{2}+2d^{2}=83$

$\implies 3\times 5^{2}+2d^{2}=83$

/* From (1) */

$\implies 75+2d^{2}=83$

$\implies 2d^{2}=83-75$

$\implies 2d^{2}=8$

$\implies d^{2}=\frac{8}{2}$

$\implies d^{2}=4$

$\implies d = \sqrt{4}$

$\implies d = ±2 ---(2)$

Therefore,

$a = 5\: and \: d = ±2$

$Case\:1: If \:a=5\:d=2\:then \\ Required \: terms \:are\\</p><p>(a-d),\:a , \:(a-d)\\</p><p>(5-2),5,(5+2)\\3,5,7\\</p><p>Case\:2 : If \:a=5 \: and \: d = -2 ,\\then \\Required \: terms \:are\\ (5+2),5,(5-2)\\=7,5,3$

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