Question

22nd sum.try and give answer

Answer 1

Hope this will help u... :-)

Answer 2

Let a & d be the first term and common difference of the given A.P respectively

a=3

d=12

a21 = a+20d

= 3+20*12

= 3+240

= 243

The required term is 120 more that the 21st term

Hence an=120+243=363

an = a+(n-1)d

363 = 3+(n-1)*12

360 = (n-1)*12

n-1 = 360/12=30

n=31

Hence the 31st term will be 120 more than the 21st term

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