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CAT 2017 Slot 2 — QA Question 34

Miscellaneous ProgressionsEasy

If a1 = 12×5, a2 = 15×8, a3 = 18×11, ......, then a1 + a2 + a3 + .... + a100 is

Answer & solution

Correct answer: 25 151

  • 25151

  • B

    12

  • C

    14

  • D

    11155

Solution

Easy

Each term has the form 1(3k−1)(3k+2)\dfrac{1}{(3k-1)(3k+2)}. Split it into partial fractions so the sum telescopes — almost everything cancels, leaving just the first and last pieces.

1

Set up the telescoping form. The denominators are 2⋅5, 5⋅8, 8⋅11,…2\cdot 5,\ 5\cdot 8,\ 8\cdot 11,\dots (gap 33). Using 1m(m+3)=13(1m−1m+3)\dfrac{1}{m(m+3)}=\dfrac13\left(\dfrac1m-\dfrac1{m+3}\right):

a1=13 ⁣(12−15),a2=13 ⁣(15−18),a3=13 ⁣(18−111), …\begin{aligned} &a_1=\tfrac13\!\left(\tfrac12-\tfrac15\right),\quad a_2=\tfrac13\!\left(\tfrac15-\tfrac18\right),\quad a_3=\tfrac13\!\left(\tfrac18-\tfrac1{11}\right),\ \dots \end{aligned}
2

Find the last term. The first factor of ana_n is 2+(n−1)⋅3=3n−12+(n-1)\cdot 3=3n-1, so for n=100n=100 it is 299299 and the second factor is 302302.

a100=13 ⁣(1299−1302)\begin{aligned} &a_{100}=\tfrac13\!\left(\tfrac1{299}-\tfrac1{302}\right) \end{aligned}
3

Telescope the sum. Adding a1a_1 through a100a_{100}, every interior fraction cancels with the next.

∑k=1100ak=13 ⁣(12−1302)(only first and last survive)⇒ ∑=13⋅151−1302⇒ ∑=13⋅150302=150906=25151\begin{aligned} &\sum_{k=1}^{100}a_k=\tfrac13\!\left(\tfrac12-\tfrac1{302}\right) \quad\text{(only first and last survive)}\\ &\Rightarrow\ \sum=\tfrac13\cdot\frac{151-1}{302}\\ &\Rightarrow\ \sum=\tfrac13\cdot\frac{150}{302}=\frac{150}{906}=\frac{25}{151} \end{aligned}
a1+a2+⋯+a100=25151a_1+a_2+\cdots+a_{100}=\dfrac{25}{151}

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CAT 2017 Slot 2 QA Q34 — Miscellaneous Progressions solution | TheCATExam