Sequences and Series 2 Question 4

4. If the sum and product of the first three terms in an AP are 33 and 1155, respectively, then a value of its 11 th term is

(2019 Main, 9 April II)

(a) 25

(b) -36

(c) -25

(d) -35

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Answer:

Correct Answer: 4. (b)

Solution:

  1. Let first three terms of an AP as ad,a,a+d.

So, 3a=33a=11

[given sum of three terms =33 and product of terms =1155]

(11d)11(11+d)=1155112d2=105d2=121105=16d=±4

So the first three terms of the AP are either 7, 11, 15 or 15,11,7.

So, the 11 th term is either 7+(10×4)=47

or 15+(10×(4))=25.

Key Idea Use the formula of sum of first n terms of AP, i.e Sn=n2[2a+(n1)d]

Given AP, is

a1,a2,a3, having sum of first n-terms

=n2[2a1+(n1)d]

[where, d is the common difference of AP]

=50n+n(n7)2A

(given)

12[2a1+(n1)d]=50+n72A

12[2a1+ndd]=5072A+n2A

a1d2+nd2=5072A+n2A On comparing corresponding term, we get

d=A and a1d2=5072Aa1A2=5072A[d=A]a1=503A So a50=a1+49d=(503A)+49A[d=A]=50+46A

Therefore, (d,a50)=(A,50+46A)



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