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Mathematics

Assertion (A) : Sum of 1st 5 terms of the G.P. : 29,13,12,....... is 5572\dfrac{2}{9}, \dfrac{1}{3}, \dfrac{1}{2}, ……. \text{ is } \dfrac{55}{72}.

Reason (R) : If for a G.P., the first term is a, the common ratio is r and number of terms = n, then sum of first n terms Sn = a(rn−1)r−1\dfrac{a(r^n - 1)}{r - 1} for all r.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

AP GP

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Answer

G.P. : 29,13,12,.......\dfrac{2}{9}, \dfrac{1}{3}, \dfrac{1}{2}, …….

Common ratio (r) : 1329=93×2=32\dfrac{\dfrac{1}{3}}{\dfrac{2}{9}} = \dfrac{9}{3 \times 2} = \dfrac{3}{2}

By formula,

Sum of G.P. (S) : a(rn−1)r−1\dfrac{a(r^n - 1)}{r - 1}

Substituting values we get :

S5=29[(32)5−1]32−1=29[(32)5−1]32−1=29[24332−1]3−22=29[243−3232]12=29×21132×2=844288=21172.S_5 = \dfrac{\dfrac{2}{9}\Big[\Big(\dfrac{3}{2}\Big)^5 - 1\Big]}{\dfrac{3}{2} - 1} \\[1em] = \dfrac{\dfrac{2}{9}\Big[\Big(\dfrac{3}{2}\Big)^5 - 1\Big]}{\dfrac{3}{2} - 1} \\[1em] = \dfrac{\dfrac{2}{9}\Big[\dfrac{243}{32} - 1\Big]}{\dfrac{3 - 2}{2}} \\[1em] = \dfrac{\dfrac{2}{9}\Big[\dfrac{243 - 32}{32}\Big]}{\dfrac{1}{2}} \\[1em] = \dfrac{2}{9} \times \dfrac{211}{32} \times 2 \\[1em] = \dfrac{844}{288} \\[1em] = \dfrac{211}{72}.

∴ Assertion (A) is false.

By formula,

Sum of G.P. (S) : a(rn−1)r−1\dfrac{a(r^n - 1)}{r - 1}

Thus,

If for a G.P., the first term is a, the common ratio is r and number of terms = n, then sum of first n terms Sn = a(rn−1)r−1\dfrac{a(r^n - 1)}{r - 1} for all r.

∴ Reason (R) is true.

Hence, Option 2 is the correct option.

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