Question

The sum of 3rd and 5th term of an ap is 82 and thee sum of its 7th and 10th terms is 46 . find the first three terms of ap

Answer 1

$Let \: \pink {a} \:and \: \blue {d}\:are \: first \:term \\ and \: Common \: difference \: of \: an \:A.P$

$We \: know \:that ,$

$\boxed { \pink { n^{th} \:term (a_{n}) = a + (n-1)d }}$

$Sum \:of \: 3^{rd} \: term \: and \: 5^{th} \:terms = 82$

$\implies a + 2d + a + 4d = 82$

$\implies 2a + 6d = 82 \: --(1)$

$Sum \:of \: 7^{th} \: term \: and \: 10^{th} \:terms = 46$

$\implies a + 6d + a + 9d = 46$

$\implies 2a + 15d = 46 \: --(2)$

/* Subtract equation (1) from equation (2), we get */

$\implies 9d = -36$

$\implies d = \frac{-36}{9}$

$\implies d = -4 \: --(3)$

/* Put d = -4 in equation (1) , we get */

$2a + 6(-4) = 82$

$\implies 2a - 24 = 82$

$\implies 2a = 82 + 24$

$\implies 2a = 106$

$\implies a = \frac{106}{2}$

$\implies a = 53 \: --(4)$

Therefore.,

$\red { First \: 3 \: terms \: of \: given \:A.P: }\\= a, ( a + d ), (a+2d ) \\= 53, [ 53 + (-4) ] \: and \: [53 + 2(-4) ] \\\green { = 53, 49 \: and \: 45 }$

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