AP ICET 2019 Shift 1 Paper

© examsiri.com

Question : 131 of 200

Marks: +1, -0

The numerically greatest term in the expansion of

(2+3x)^5

when x =

5/4

is

${3×15^2}/8$
${5×15^2}/8$
${5×15^4}/{16}$
${3×15^4}/{16}$

Solution:

Given

(2 + 3x)^5

=2^5(1+3/2x)^5

Given

x=5/4

|\;{3}/{2} x |=\;{3}/{2} × \;{5}/{4}=\;{15}/{8}

and

n =5

Let

P =\;{( n +1)| x |}/{| x |+1}

=\;{6 × {15}/{8}}/{{15}/{8}+1}=\;{90}/{23}

[P]=3

∴T_{[P]+1}=T_4

is the numerically greatest term.

T _{4}=5 c _{3}(2)^{5-3}(\;{15}/{4})^{3}

=10\;\;2^{2} \;{15^{3}}/{4^{3}}

=\;{5.15^{3}}/{8}

© examsiri.com

Go to Question:

Prev Question

Next Question