Sequence , Series and Logarithm Question no. 1 What is the eighth term of the sequence 1, 4, 9, 16, 25.... ? looks_one 8 looks_two 128 looks_3 64 looks_4 None of these Answer Option looks_3 64 Solution Solution : Given series consists of the perfect square terms , so given sequence is 1 , 4 ,9 , 16 ,25 , 36 , 49 , 64 ..... so the 8th term is 64 Question no. 2 The value of (1 3 +2 3 +3 3 +....+15 3 ) – (1+2+3+.... 15) is: looks_one 14280 looks_two 14400 looks_3 12280 looks_4 None of these Answer Option looks_one 14280 Solution Solution : For cubic power terms (1 3 +2 3 +3 3 +....+15 3 ) – (1+2+3+.... 15) = 14400 - 120 = 14280 Question no. 3 Iflog a b= 1/2, log b c= 1/3 and log c a = K/5 , then the value of K looks_one 25 looks_two 35 looks_3 20 looks_4 30 Answer option looks_4 30 Solution Solution : log a b = 1/2 log b c = 1/3 log a c = K/5 log a b * lo...

Progression Question no. 31 If a,b,c are distinct such that ab+bc+ca≠0 and in AP, then a 2 (b+c) , b 2 (a+c) , c 2 (b+a) is looks_one AP looks_two GP looks_3 HP looks_4 none of these Answer Option looks_one AP Solution Solution : a ,b ,c are in A.P. 2 (b+c) , b 2 (a+c) , c 2 (b+a) is" border="0" width="400" data-original-height="684" data-original-width="1654" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgodfK_UWYlYD8RqsSqmNKRyipcDz9b5ABqT3If2lyUKT5TrDgbFkfcMFcuI1jba1B_k1D3KPF7-kOOUkvptjmXwoniVL8SGXUxKMpyimKDw_lovHJYG_AQa8Os4W4Kq45PgNSCfnhq9ds/s400/progression_solution_31_Maths_queston_IITjee.png"/> Question no. 32 The sum of the series: 1 2 + 3 2 + 5 2 + .......... n terms is: looks_one (4n 2 -1) looks_two (4n-1) looks_3 (4n+1) looks_4 (4n 2 +1) Answer Option looks_one (4n 2 -1) solution Solution : Question no. 33 ...

Progression Question no. 16 The sum of 24 terms of the series √ 2 + √ 8 + √ 18 + √ 32 .... is looks_one 300 looks_two 300√ 2 looks_3 200√ 2 looks_4 none of these Answer Option looks_two 300√ 2 is correct answer Solution Given series is ,√ 2 + √ 8 + √ 18 + √ 32 ⇒ it can be write down like √ 2 + 2√ 2 + 3√ 2 + 4√ 2 ,.... ⇒ it is clearly observed that given series is in A.P. with ⇒ a= √ 2 d =√ 2 ⇒ sum of first 24 terms = (n/2) { 2a+ (n-1)d } = (24/2) { 2√ 2 + (24-1)√ 2 } = 12 { 2√ 2 +23√ 2 } = 12 { 25√ 2 } = 300√ 2 Question no. 17 If log 3 2, log 3 (2 x -5) and log 3 (2 x -7/2) are in A.P. , then the value of x is looks_one 1, 1/2 looks_two 1, 1/3 looks_3 1, 3/2 looks_4 none of these Answer option looks_...