which term of the AP 3,14,25,36..... will be 99 more than its 25 th term .

soln,
given,
a=3 and d=11
now , according to question ,
                                            an=a25+99
                                      or,  a+(n-1)d=a+24.d+99
                                      or,3+(n-1)11=3+24*11+99
                                      or, 3+11n-11=264+3+99
                                      or, 11n-8=366
                                     or, 11n=366+8
                                    or ,  11n=374
                                     or, n=374/11=34
therefore, the required term= 34th term

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hi buddy!!!
wanna know the answer????
eeeeeeeeeeeeee

let A be the first term and D be the common difference.
now 1ST TERM ,a = 3
and DIFFERENCE,d = 11
let nth term of an AP is 99 more than 25th term
so an = a25 + 99
a+(n-1)d = a + (n-1)d +99
3 + (n-1)11 = 3 + (25 -1) 11 +99
3+ (11n-11) = 3 + (24)(11)+99
11n-11= 99+264
11n = 99 + 264+ 11
11n = 374
n = 374/11
n = 34
therefore, 34th term will be 99 more than its 25th term
prob. solved buddy!!!!!!! 



 
 
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