Which term of the A . P . 3 , 10 , 17 , … will be 84 more than its 13 th term?

A . P . is 3 , 10 , 17 , … . Here, First term ( a ) = 3 Common difference of the A . P . ( d ) = 10 - 3 = 7 Now, as we know, a n = a + ( n - 1 ) d So, for 13 t h term ( n = 13 ) , a 13 = 3 + ( 13 - 1 ) ( 7 ) = 3 + 12 ( 7 ) = 3 + 84 = 87 Let us take the term which is 84 more than the 13 t h term as a m . So, a m = 84 + a 13 = 84 + 87 = 171 Also, a m = a + ( m - 1 ) d 171 = 3 + ( m - 1 ) 7 171 = 3 + 7 m - 7 171 = - 4 + 7 m 171 + 4 = 7 m Further simplifying, we get : 175 = 7 m m = 175 7 m = 25 Therefore, the required answer is 25 t h term.

Which term of the arithmetic progression 8 , 14 , 20 , 26 , … will be 72 more than its 41 st term.

In the given problem, let us first find the 41 st 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 41 st term, a 41 = 8 + ( 41 - 1 ) ( 6 ) = 8 + 40 ( 6 ) = 8 + 240 = 248 Let us take the term which is 72 more than the 41 st term as a n . So, a n = 72 + a 41 = 72 + 248 = 320 Also, a n = a + ( n - 1 ) d 320 = 8 + ( n - 1 ) 6 320 = 8 + 6 n - 6 320 = 2 + 6 n 320 - 2 = 6 n Further simplifying, we get, 318 = 6 n ⇒ n = 318 6 ⇒ n = 53 Therefore, the 53 rd term of the given A.P. is 72 more than the 41 st term.