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JEE Mathematics 2024 Question with Solution

If in a G.P. of 6464 terms, the sum of all terms is 77 times the sum of the odd terms, then the common ratio of the G.P. is equal to:

  • A

    77

  • B

    44

  • C

    55

  • D

    66

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A G.P. has 6464 terms, and the sum of all terms is 77 times the sum of the odd terms.

Find: The common ratio rr.

Let the first term be aa and the common ratio be rr.

The sum of all 6464 terms is

S64=a(r64−1)r−1S_{64} = \frac{a(r^{64}-1)}{r-1}

The odd terms are

a,ar2,ar4,…,ar62a, ar^2, ar^4, \dots, ar^{62}

which form a G.P. with first term aa, common ratio r2r^2, and 3232 terms.

So, their sum is

Sodd=a(r64−1)r2−1S_{\text{odd}} = \frac{a(r^{64}-1)}{r^2-1}

According to the question,

a(r64−1)r−1=7⋅a(r64−1)r2−1\frac{a(r^{64}-1)}{r-1} = 7 \cdot \frac{a(r^{64}-1)}{r^2-1}

Canceling the common factor a(r64−1)a(r^{64}-1), we get

1r−1=7r2−1\frac{1}{r-1} = \frac{7}{r^2-1}

Cross-multiplying,

r2−1=7(r−1)r^2 - 1 = 7(r - 1)

so

r2−1=7r−7r^2 - 1 = 7r - 7

and hence

r2−7r+6=0r^2 - 7r + 6 = 0

Factoring,

(r−1)(r−6)=0(r-1)(r-6)=0

Thus,

r=1 or r=6r=1 \text{ or } r=6

From the solution, r=1r=1 is discarded, so the common ratio is 66.

Therefore, the correct option is D.

Using explicit GP terms

Given: The G.P. is

a,ar,ar2,ar3,…,ar63a, ar, ar^2, ar^3, \dots, ar^{63}

and the total sum is 77 times the sum of the odd terms.

Find: The common ratio rr.

The total sum is

S=a+ar+ar2+⋯+ar63=a(1−r64)1−rS = a + ar + ar^2 + \dots + ar^{63} = \frac{a(1-r^{64})}{1-r}

The odd-position terms are

a+ar2+ar4+⋯+ar62a + ar^2 + ar^4 + \dots + ar^{62}

So,

Sodd=a(1−r64)1−r2S_{\text{odd}} = \frac{a(1-r^{64})}{1-r^2}

Given that

S=7SoddS = 7S_{\text{odd}}

we get

a(1−r64)1−r=7⋅a(1−r64)1−r2\frac{a(1-r^{64})}{1-r} = 7\cdot \frac{a(1-r^{64})}{1-r^2}

Canceling a(1−r64)a(1-r^{64}),

11−r=71−r2\frac{1}{1-r} = \frac{7}{1-r^2}

Cross-multiplying,

1−r2=7(1−r)1-r^2 = 7(1-r)

which simplifies to

r2−7r+6=0r^2 - 7r + 6 = 0

Factorizing,

(r−6)(r−1)=0(r-6)(r-1)=0

Hence,

r=6 or r=1r=6 \text{ or } r=1

The solution concludes that r=1r=1 does not satisfy the intended condition, so we take r=6r=6.

Therefore, the correct option is D.

Common mistakes

  • Using the sum of odd terms with common ratio rr instead of r2r^2 is incorrect because the odd-position terms are a,ar2,ar4,…a, ar^2, ar^4, \dots. Always identify the new G.P. formed by the selected terms before applying the sum formula.

  • Taking the number of odd terms as 6464 instead of 3232 is wrong. In 6464 total terms, exactly half are in odd positions, so the odd-term sum must use 3232 terms.

  • Canceling factors carelessly and writing the wrong simplified equation can lead to an incorrect quadratic. After substitution, cancel only the common factor a(r64−1)a(r^{64}-1) and then cross-multiply carefully.

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