If the third term of a GP is 4, then the product of first five terms of the GP is<\/p>\n
[amp_mcq option1=”43<\/sup>” option2=”44<\/sup>” option3=”45<\/sup>” option4=”46<\/sup>” correct=”option3″]<\/p>\n We are given that the third term of the GP is 4. We need to find the product of the first five terms of the GP. Let’s group the ‘a’ terms and the ‘r’ terms: We can rewrite this expression by grouping $(ar^2)$ terms:
\nLet the first term of the GP be ‘a’ and the common ratio be ‘r’.
\nThe terms of the GP are:
\n1st term: $T_1 = a$
\n2nd term: $T_2 = ar$
\n3rd term: $T_3 = ar^2$
\n4th term: $T_4 = ar^3$
\n5th term: $T_5 = ar^4$<\/p>\n
\n$T_3 = ar^2 = 4$.<\/p>\n
\nProduct P = $T_1 \\times T_2 \\times T_3 \\times T_4 \\times T_5$
\nP = $a \\times (ar) \\times (ar^2) \\times (ar^3) \\times (ar^4)$<\/p>\n
\nP = $(a \\times a \\times a \\times a \\times a) \\times (r^0 \\times r^1 \\times r^2 \\times r^3 \\times r^4)$
\nP = $a^5 \\times r^{(0+1+2+3+4)}$
\nP = $a^5 \\times r^{10}$<\/p>\n
\nP = $a^5 \\times r^{10} = (a^5 \\times r^{10\/5 \\times 5}) = (a^5 \\times r^{2 \\times 5})$
\nP = $(a \\times r^2)^5$
\nP = $(ar^2)^5$<\/p>\n