Question

Find 3 consecutive terms in an A.P. whose sum is 18 and thier squares is 140.

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Answer 1

Answer:

Step-by-step explanation:

Answer 2

     

$\sf{Let the consecutive terms be} \ a-d,\ a,\ a+d.$

$(a-d)+a+(a+d)=18 \\ \\ a-d+a+a+d=18 \\ \\ 3a=18 \\ \\ a=6$

$(a-d)^2+a^2+(a+d)^2=140 \\ \\ (6-d)^2+36+(6+d)^2=140 \\ \\ (6-d)^2+(6+d)^2=140-36 \\ \\ (6-d)^2+(6+d)^2=104 \\ \\ 2(36+d^2)=104 \ \ \ \ \ [(a-b)^2+(a+b)^2=2(a^2+b^2)] \\ \\ 36+d^2=52 \\ \\ d^2=16 \\ \\ d = \pm 4$

$a-d=6-4=2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a-d=6+4=10 \\ \\ a=6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{ OR }\ \ \ \ \ \ \ \ \ \ \ \ \ \ a=6 \\ \\ a+d=6+4=10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a+d=6-4=2$

$\therefore\ \sf{2, 6 and 10 are the terms. \\ \\ \\ Plz mark it as the brainliest. \\ \\ \\ Thank you. :-)}$