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The Sixth Term Of An Ap Is Equal To 2 And Its Common Difference Is Greater Than 1

(1) -8/5 (2) 8/5 (3) -5/8 (4) 5/8 Solution: Let a be the first term and d be the common difference of the AP. Given 6th term = 2... View Article

(1) 50 (2) 75 (3) 100 (4) 125 Solution: Given sequence 1, 3, 6, 10, 15,…., 5050 Let n be the number of terms. Sn =... View Article

(1) n/2 (2) -n/2 (3) n/3 (4) -n/3 Solution: Given m, n, r are in HP. So 2/n = (r+m)/mr mr = n(m+r)/2 (m+r) = 2mr/n…(i) Also (m... View Article

(1) 11e-4 (2) 11e-6 (3) 10e+5 (4) 3e+4 Solution: 9+16/2! + 27/2! + 42/4! +....∞ Here Tn = (2n2+n+6)/n! Sn = Σ(2n2+n+6)/n! =... View Article

(1) AP (2) GP (3) HP (4) None of these Solution: We know tan (A+B) = (tan A+ tan B) /(1-tan A tan B) tan 700 = tan (500+ 200)... View Article

(1) e2 (2) loge 2 +1 (3) loge 3 -2 (4) 1-loge 2 Solution: We know logb c = 1/logc b Let S = log4 2 - log8 2 + log16 2 +.. =... View Article

(1) 4e-1 (2) 4e-3 (3) 3e+2 (4) 3e+4 Solution: Given 3/1! + 5/2! + 9/3! + 15/4 ! + 23/5! +.....∞ use e = 1+ 1/1! +1/2! + 1/3! +... View Article

(1) e (2) 2e (3) 3e (4) none of these Solution: Given series 2/3! + 4/5! + 6/7! +.....to ∞ Tn = 2n/(2n+1)! = (1/2n!) -... View Article

(1) e (2) 2e (3) 3e (4) none of these Solution: 1.2/1! + 2.3/2! + 3.4/3! +....∞ Tn = n(n+1)/n! = (n+1)/(n-1)! Put n = 1,2,3,...... View Article

(1) 33/6 (2) (log 3)3/3 (3) 3/3! (4) None of these Solution: We know ax = 1+ (x ln a)/1! + (x ln a)2/2! + (x ln a)3/3! + ….. 3x =... View Article

(1) 1+ 22x2/2! + 24x4/4! + … (2) 1+(2x)2/2! + 22x4/4! + … (3) 1+(2x)2/2.2! + 2x4/4! + … (4) 1+(2x)2/2.2! + (2x)4/2.4! +... Solution:... View Article

(1) (ab)n/n! (2) eban/n! (3) ea.bn/n! (4) ea+b (ab)n/n! Solution: We know ex = 1 + x/1! + x2/2! + x3/3! +... 1+(a+bx)/1! +... View Article

(1) e (2) e-1 (3) 1 (4) 0 Solution: (1/e3x )(ex + e5x) = a0 + a1x + a2x2 + … e-2x+e2x = a0 + a1x + a2x2 + … Put x = 2 e-4+e4 =... View Article

(1) loge (4/3) (2) loge (3/4) (3) 2loge (3/4) (4) 2loge (4/3) Solution: 2[7-1 + 3-1 . 7-3 + 5-1 . 7-5 +...] = 2[1/7 + ⅓. 1/73 + ⅕... View Article

(1) loge 2 – 1/2 (2) loge 2 (3) loge 2 + 1/2 (4) loge 2 +1 Solution: S = 1/1.2.3 + 1/3.4.5 + 1/5.6.7 + …. = 1/(2n-1)(2n)(2n+1) =... View Article

(1) n/a1an (2) (n-1)/a1an (3) (n+1)/a1an (4) 2n/a1an Solution: = = = = = = = Hence option (2) is the answer.... View Article

(1) 1020300/3 (2) 1030200/3 (3) 1003200/3 (4) 1023200/3 Solution: 2 + 6 + 12 + 20 + 30 + ... = (12+1) + (22+2) + (32+3)+...upto... View Article

(1) e2 (2) loge 2 + 1 (3) loge 3 - 2 (4) 1 - loge 2 Solution: Given S = log4 2 - log8 2 + log16 2 - … S = ½ - ⅓+ ¼-.... We use... View Article

(1) log(1- px + qx2 ) (2) log(1 + px - qx2 ) (3) log(1 + px + qx2 ) (4) None of these Solution: Given α ,β are the roots of the... View Article

(1) 2-n (2) 2-n(n - 1) (3) 2n (n - 1) + 1 (4) 2-n + n - 1 Solution: S = 1/2 + 3/4 + 7/8 + 15/16 + …..n terms = (2-1)/2 + (4-1)/4... View Article