CAT 2018 Question Paper | Quants Slot 2

CAT Previous Year Paper | CAT Sequence and Series Questions | Question 5

In this question, you are asked to find the value of k, given a sequence with conditions. This question appeared in the CAT 2018 Question Paper Slot 2. If youa re looking to ace you CAT online preparation, solving CAT Previous Year Paper is a must for you! Start with this question right away!

Question 5 :Let t1, t2,.....be real numbers such that t1+ t2 +... + tn = 2n2 + 9n + 13, for every positive integer n ≥ 2. If tk=103, then k equals[TITA]


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Explanatory Answer

Method of solving this CAT Question from Sequence and Series

Let t1, t2,.....be real numbers such that t1+ t2 +... + tn = 2n2 + 9n + 13
If tk = 103 we have to find the value of k.
Let us assume this as sum to n terms i.e.
Sn = t1+ t2 +... + tn = 2n2 + 9n + 13
Given an expression in terms of n we can also take sum to n-1 terms i.e.
Sn-1 = t1+ t2 +... + tn-1 = 2(n-1)2 + 9(n-1) +13
This Sn and Sn-1 can be subtracted one from the another such that,
Sn = t1+ t2 +... + tn = 2n2 + 9n + 13
Sn-1 = t1+ t2 +... + tn-1 = 2(n-1)2 + 9(n-1) +13
Sn - Sn-1 = tn = 2n2 + 9n + 13 - 2(n-1)2 - 9(n-1) - 13
Sn - Sn-1 = tn = 2n2 + 9n + 13 - 2(n-1)2 - 9n + 9 - 13
Sn - Sn-1 = tn = 2[n2 - (n-1)2] + 9
Sn - Sn-1 = tn = 2[n2 - n2 + 2n - 1] + 9

tn = 2[2n - 1] + 9
tn = 4n - 2 + 9
tn = 4n + 7
So tk = 4k + 7 = 103
k = \\frac{103 - 7}{4})
k = 24
The 24th term is 103

The question is "Let t1, t2,.....be real numbers such that t1+ t2 +... + tn = 2n2 + 9n + 13, for every positive integer n ≥ 2. If tk=103, then k equals [TITA] "

Hence, the answer is 24

 

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