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NCERT Exemplar · Q18

Q.The third term of G.P. is 44. The product of its first 55 terms is
(A) 434^3
(B) 444^4
(C) 454^5
(D) None of these

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For a Geometric Progression with an odd number of terms, the product of these terms is equal to the middle term raised to the power of the number of terms. Given the third term is 44 and there are 55 terms, the product is 454^5.

A Geometric Progression (G.P.) is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. If the first term is aa and the common ratio is rr, the terms are a,ar,ar2,ar3,…a, ar, ar^2, ar^3, \dots.

When we need to find the product of terms in a G.P., especially an odd number of terms, there's a very elegant property we can use. Consider a G.P. with nn terms. If nn is odd, there is a unique middle term.

Why this approach works:

Let the five terms of the G.P. be a1,a2,a3,a4,a5a_1, a_2, a_3, a_4, a_5.

In a G.P., there's a symmetrical relationship between terms equidistant from the beginning and the end.

Specifically, a1⋅a5=a2⋅a4=a3⋅a3=a32a_1 \cdot a_5 = a_2 \cdot a_4 = a_3 \cdot a_3 = a_3^2.

This is because if the terms are a,ar,ar2,ar3,ar4a, ar, ar^2, ar^3, ar^4:

a1⋅a5=a⋅(ar4)=a2r4a_1 \cdot a_5 = a \cdot (ar^4) = a^2 r^4

a2⋅a4=(ar)⋅(ar3)=a2r4a_2 \cdot a_4 = (ar) \cdot (ar^3) = a^2 r^4

a3⋅a3=(ar2)⋅(ar2)=a2r4a_3 \cdot a_3 = (ar^2) \cdot (ar^2) = a^2 r^4

So, the product of the first five terms is P=a1⋅a2⋅a3⋅a4⋅a5P = a_1 \cdot a_2 \cdot a_3 \cdot a_4 \cdot a_5.

We can group these terms: P=(a1⋅a5)⋅(a2⋅a4)⋅a3P = (a_1 \cdot a_5) \cdot (a_2 \cdot a_4) \cdot a_3.

Using the property above, we substitute: P=(a32)⋅(a32)⋅a3=a32+2+1=a35P = (a_3^2) \cdot (a_3^2) \cdot a_3 = a_3^{2+2+1} = a_3^5.

This means the product of the first 55 terms is simply the third term raised to the power of 55. This generalises: for nn terms, the product is (middle term)n^n.

Now, let's apply this to the given problem.

  1. Represent the terms of the G.P.:

    Let the first term of the G.P. be aa and the common ratio be rr.

    The first five terms are:

    T1=aT_1 = a

    T2=arT_2 = ar

    T3=ar2T_3 = ar^2

    T4=ar3T_4 = ar^3

    T5=ar4T_5 = ar^4

  2. Use the given information:

    We are given that the third term of the G.P. is 44.

    So, T3=ar2=4T_3 = ar^2 = 4.

  3. Calculate the product of the first 55 terms:

    The product PP of the first 55 terms is:

    P=T1⋅T2⋅T3⋅T4⋅T5P = T_1 \cdot T_2 \cdot T_3 \cdot T_4 \cdot T_5

    P=(a)⋅(ar)⋅(ar2)⋅(ar3)⋅(ar4)P = (a) \cdot (ar) \cdot (ar^2) \cdot (ar^3) \cdot (ar^4) …

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