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Which Term of the A.P. 8, 14, 20, 26, ... Will Be 72 More than Its 41st Term? - CBSE Class 10 - Mathematics

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Question

Which term of the A.P. 8, 14, 20, 26, ... will be 72 more than its 41st term?

Solution

In the given problem, let us first find the 41st term of the given A.P.

A.P. is 8, 14, 20, 26 …

Here,

First term (a) = 8

Common difference of the A.P. (d) = 14 - 8 = 6

Now as we know

`a_n = a+(n + 1)d`

So for 41st term (n = 41)

`a_(41) = 8 + (41 - 1)(6)`

= 8 + 40(6)

= 8 + 240

= 248

Let us take the ter which is 72 more than the 41st term as an  So

`a_n = 72 + a_41`

= 72 + 248

= 320

Also `a_n = a + (n - 1)d`

320 = 8 + (n - 1)6

320 = 8 + 6n - 6

320 = 2 + 6n

320 - 2 = 6n

Futher simplifying weget

318 = 6n

`n = 318/6`

n = 53

Therefore the 53rd term of the given A.P is 72 more than the the 41st term.

  Is there an error in this question or solution?

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Solution Which Term of the A.P. 8, 14, 20, 26, ... Will Be 72 More than Its 41st Term? Concept: nth Term of an AP.
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