Question

which term of ap:3,8,13,18......will be 130 more than its 31st term

Answer 1

Hey mate..
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Given ,

AP:-- 3,8,13,18.......

Here ,

First term, a = 3

Common Difference , d = ( 8-3 ) = 5

a(n) = ?

We know,

a(n) = a + ( n - 1 ) × d

Let n th term of the AP be 130 more than its 31st term

Therefore,

=> a(n) = a(31) + 130

=> a + ( n -1 ) d = a + 30d + 130

=> 3 + ( n - 1 ) 5 = 3 + 30×5 + 130

=> 5n - 2 = 3 + 150 + 130

=> 5n = 285

=> n = 285 / 5

=> n = 57

Therefore,

Required n th term is 57

#racks.

Answer 2

Hello Friend ✋✋✋

Here ,

a = 3

Common difference , d = 5

$so \: we \: need \: to \: find \: n \: when \: a_{n} = a_{31} + 130$
$a_{31} = 3 + 30 \times 5 \\ = 153$
$a_{n} = a_{31} + 130 \\ = 153 + 130 = 283 \:$
$by \: using \: a_{n} = a + (n - 1)d \: to \: find \: a_{n} \\ 283 = 3 + (n - 1) \times 5 \\ 280 \div 5 = n - 1 \\ 56 + 1 = n \\ n = 57$

So your answer is 57....

Hope it helps