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CAT 2018 Slot 2 — QA Question 32

Miscellaneous ProgressionsEasy

The value of the sum 7 × 11 + 11 × 15 + 15 × 19 + ...+ 95 × 99 is

Answer & solution

  • A

    80730

  • B

    80773

  • C

    80751

  • 80707

Solution

Easy

Each term is a product of two numbers from arithmetic progressions with common difference 44. Write the general term, count the terms, then sum using the standard formulas ∑n2=n(n+1)(2n+1)6\sum n^2=\tfrac{n(n+1)(2n+1)}{6} and ∑n=n(n+1)2\sum n=\tfrac{n(n+1)}{2}.

1

Identify the general term. First factors 7,11,15,…,957,11,15,\dots,95 give 3+4n3+4n; second factors 11,15,19,…,9911,15,19,\dots,99 give 7+4n7+4n (for n=1,…,23n=1,\dots,23).

number of terms=95−74+1=23⇒ Tn=(3+4n)(7+4n)⇒ Tn=16n2+40n+21\begin{aligned} &\text{number of terms} = \frac{95-7}{4} + 1 = 23\\ &\Rightarrow\ T_n = (3+4n)(7+4n)\\ &\Rightarrow\ T_n = 16n^2 + 40n + 21 \end{aligned}
2

Sum term by term over n=1n=1 to 2323. Split the sum.

∑n=123Tn=16∑n2+40∑n+21∑1⇒ ∑n=123n2=23⋅24⋅476=4324⇒ ∑n=123n=23⋅242=276\begin{aligned} &\sum_{n=1}^{23} T_n = 16\sum n^2 + 40\sum n + 21\sum 1\\ &\Rightarrow\ \sum_{n=1}^{23} n^2 = \frac{23\cdot 24\cdot 47}{6} = 4324\\ &\Rightarrow\ \sum_{n=1}^{23} n = \frac{23\cdot 24}{2} = 276 \end{aligned}
3

Combine. Plug the partial sums from step 2.

16(4324)+40(276)+21(23)⇒ =69184+11040+483⇒ =80707\begin{aligned} &16(4324) + 40(276) + 21(23)\\ &\Rightarrow\ = 69184 + 11040 + 483\\ &\Rightarrow\ = 80707 \end{aligned}
80707\textbf{80707}
CAT 2018 Slot 2 QA Q32: The value of the sum 7 × 11 + 11 × 15 + 15 × 19 + ...+ 95 × 99 is — Solution | TheCATExam