JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Medium

1 of 1

Application +4 | -1

If 7 th and 13 th terms of an AP are 34 and 64 respectively, then its 18 th term is

View answer and explanation

Correct option: C

Option C (89) is correct because it is the 18th term of the AP, calculated using the given 7th and 13th terms to find the common difference and first term.

More Questions from SEQUENCE OF SERIES

Practice All Chapter MCQs →

Given the sets

A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

, then find the following
(i)

{A} \cup({B} \cap {C})

(ii)

A-(B \cap C)

(iii)

(B \cup C)-A

a,b,c

are in A.P.;

a,x,b

are in G.P. and

b,y,c

are in G.P., then

{x^2},{b^2},{y^2}

are in Suppose

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

is expressed as a polynomial in

y

, as

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

where

y=x-2

, then

\sum\limits_{{i=0}}^{2012} a_i

equals If

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

and

\log \left(\frac{a}{3 b}\right)

are in

A P

, where

a, b

and c are in

G P

, then

a, b, c

are the lengths of sides of If the sum of three numbers in G.P. is 63 and the product of the first and the second term is

\frac{3}{4}

of the third term, then the numbers are If the lengths of sides of a right angled triangle are in A.P., then their ratio is