Question

Hi



Which term of an ap 8, 14, 20, 26...........Will be 72 more than its
${41}^{st}$
Term?

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Answer 1

So According to the question

It is a A.P

here

a = 8

d = 14 - 8

= 6

its 41st term will be = a + 40 d

= 8 + 40 × 6

= 8 + 240

= 248

Now we have to find the term which is 72 more than 41st term so 72 + 248

= 320

so by formula

an = a + ( n - 1) d

320 = 8 + (n - 1) 6

320 = 8 + 6n - 6

320 = 2 + 6n

320 - 2 = 6n

318 = 6n

n = 318/6

n = 53

Answer

so 53th term

Answer 2

$\underline\bold{\huge{GOOD\: AFTERNOON\: DEAR\: :}}$

Given:a=8

d=14-8=6

Let 'n' term be 72 more than its 41 term.

So, a+(n-1)d=a+(41-1)d+72

a+(n-1)d=a+40d+72

(n-1)d=40d+72 (a got cancelled)

=>(n-1)6=40d+72

putting the value we got

=>6n-6=40×6+72

=>6n=240+72-6

=>6n=240+66

=>6n=306

=>n=306÷6

=>n=51

$\underline\bold{So, The\: term\: is\: 51 \:terms.\: :}$