PYQSide
ExamsMy practice
Consider an A.P.: a1, a2, , an; a1 > 0. If a2 - a1 = -34, an = 14a1, and i=1n ai = 5252, then i=117 ai is equal to
(1) 238
Common difference: d = a2 - a1 = - 34
an = a1 + (n-1)d = 14a1
34a1 = 34(n-1) a1 = n-1
i=1n ai = n2(a1 + an) = n2 54a1 = 5n(n-1)8
5n(n-1)8 = 5252
n(n-1) = 420 = 21 20
n = 21, a1 = 20
i=117 ai = 172 [2(20) + 16 (- 34 ) ]
= 172(40 - 12) = 17 14 = 238
Answer: (1)
Let a1, a2, a3, a4 be an A.P. of four terms such that each term of the A.P. and its common difference l are integers. If a1 + a2 + a3 + a4 = 48 and a1a2a3a4 + l4 = 361, then the largest term of the A.P. is equal to
( 13 + 47 ) + ( 132 + 13 47 + 4272 ) + ( 133 + 132 47 + 13 4272 + 4373 ) + up to infinite terms, is equal to
Let k=1n ak = n2 + n. If a10 = 59 and a6 = 7a1, then + is equal to :
If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is
Suppose a, b, c are in A.P. and a2, 2b2, c2 are in G.P. If a < b < c and a + b + c = 1, then 9(a2 + b2 + c2) is equal to .
Let a1, a2, a3, be a G.P. of increasing positive terms such that a2 a3 a4 = 64 and a1 + a3 + a5 = 8137. Then a3 + a5 + a7 is equal to :
Let a1 = 1 and for n 1, an+1 = 12an + n2 - 2n - 1n2(n + 1)2. Then | n=1 (an - 2n2 ) | is equal to .
The positive integer n, for which the solutions of the equation x(x + 2) + (x + 2)(x + 4) + + (x + 2n - 2)(x + 2n) = 8n3 are two consecutive even integers, is :
Let a1, a22, a322, , a1029 be a G.P. of common ratio 1 2. If a1 + a2 + + a10 = 62, then a1 is equal to:
If 114 + 124 + 134 + = 490, 114 + 134 + 154 + = , 124 + 144 + 164 + = , then is equal to :