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Sequences and Series question

2023 · 10 Apr · Shift 1 · Q37
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Sequences and Series question

2023 · 10 Apr · Shift 1 · Q37

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let the first term α\alphaα and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to
  1. A
    241
  2. B
    231
  3. C
    220
  4. D
    210
View written solutionFree

Correct answer: B

  1. Write the first three terms of the GP

If the first term is α\alphaα and common ratio is rrr, then the first three terms are

α,  αr,  αr2\alpha,\; \alpha r,\; \alpha r^2α,αr,αr2

where α,r\alpha, rα,r are positive integers.

  1. Use the condition on the sum of squares

Given:

α2+(αr)2+(αr2)2=33033\alpha^2 + (\alpha r)^2 + (\alpha r^2)^2 = 33033α2+(αr)2+(αr2)2=33033

So,

α2(1+r2+r4)=33033\alpha^2(1+r^2+r^4)=33033α2(1+r2+r4)=33033
  1. Factorize 330333303333033

Let us factorize:

33033=3×11011=3×7×1573=3×7×11×14333033=3\times 11011=3\times 7\times 1573=3\times 7\times 11\times 14333033=3×11011=3×7×1573=3×7×11×143

Also,

143=11×13143=11\times 13143=11×13

Hence,

33033=3×7×112×1333033=3\times 7\times 11^2\times 1333033=3×7×112×13

So,

33033=1812+?33033=181^2+? 33033=1812+?

But instead of guessing, we use

α2(1+r2+r4)=33033.\alpha^2(1+r^2+r^4)=33033.α2(1+r2+r4)=33033.

Thus α2\alpha^2α2 must divide 330333303333033.

  1. Find possible values of α\alphaα

Since

33033=3×7×112×13,33033=3\times 7\times 11^2\times 13,33033=3×7×112×13,

the only square factor greater than 111 is

112=121.11^2=121.112=121.

So possible integer values of α\alphaα are

α=1orα=11.\alpha=1 \quad \text{or} \quad \alpha=11.α=1orα=11.
  1. Test α=11\alpha=11α=11

Then

1+r2+r4=33033121=273.1+r^2+r^4=\frac{33033}{121}=273.1+r2+r4=12133033​=273.

Let x=r2x=r^2x=r2. Then

x2+x+1=273x^2+x+1=273x2+x+1=273

so

x2+x−272=0.x^2+x-272=0.x2+x−272=0.

Now,

172+17−272=289+17−272=34≠0,17^2+17-272=289+17-272=34\neq 0,172+17−272=289+17−272=34=0, 162+16−272=256+16−272=0.16^2+16-272=256+16-272=0.162+16−272=256+16−272=0.

Thus

x=16⇒r2=16⇒r=4x=16 \Rightarrow r^2=16 \Rightarrow r=4x=16⇒r2=16⇒r=4

(since rrr is a positive integer).

Hence the three terms are

11,  44,  176.11,\;44,\;176.11,44,176.

Their sum is

11+44+176=231.11+44+176=231.11+44+176=231.
  1. Test α=1\alpha=1α=1

Then

1+r2+r4=33033.1+r^2+r^4=33033.1+r2+r4=33033.

Let x=r2x=r^2x=r2. Then

x2+x+1=33033x^2+x+1=33033x2+x+1=33033

which gives

x2+x−33032=0.x^2+x-33032=0.x2+x−33032=0.

Its positive integer solution would need discriminant

1+4(33032)=1321291+4(33032)=1321291+4(33032)=132129

to be a perfect square, but it is not. So no integer rrr arises.

Therefore the only valid GP is

11,  44,  176.11,\;44,\;176.11,44,176.
  1. Final answer

The sum of the three terms is

231\boxed{231}231​

So the correct option is B.

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