Question

find 99th term of an AP with series -10,-6,-2
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Answer 1

$\huge\boxed{Answer:-}$

$\bf{Given:-}$

Ap is -10 , -6 , -2 .......

so , 1st term [a] is = -10

2nd term [ a2 ] = -6

Now , Common difference [ d ] = { a2 - a1 }

{ -10 - ( -6)} = { -10 + 6 }= - 4

so,$\bf{99th \:term \:of \:an \:is \:an = a99}$

so a99 = a + ( n - 1 ) d

⇒ a99 = -10 + ( 99 - 1 ) -4

⇒ a99 = -10 + [ 98 × ( -4 )]

⇒ a99 = -10 - 392

⇒ a99 = - 402

so, the 99th term of an Ap is -402