explain the whole solution especially that part in which we put n=23 we do we do that

explain the whole solution especially that part in which we put n=23 we do we do that SEQUENCES AND SERIES 183 or or em find thei Now (3) (4) 2 711+15 U[2a2 +(n —1)d2] 2 24 _ 311+8 2a2 711+15 12th term of first A.P. 12th term of second A.P +1 ld2 24 +224 3 x 23+8 [By putting n = 23 in (1)1 +22d2 703+15 12th term of first A.P. 7 +114 Therefore +11612 12th term of second A.P. 16 Hence, the required ratio is 7 : 16. 0. •n the first year and he receives

Dear student,

In first step we put the formula of sum of for two APs.
In second step both 2 get cancelled.

And formula of ​ terms for two APs.
We take  = 23 to compare with equation (1), so that both become exact same.
This is HIT and TRIAL method for solving these type of questions.

Regards
 

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