Let a_1,a_2,,a_30 be an A.P., S = _i = 1^302mu a_i and T = _i = 1^152mu a_(2i - 1). If a_5 = 27 and S - 2T = 75, then a_10 is equal to

(b) : Let d is the common difference of the A.P.aligned& S=_i=1^30 a_i=302[a_1+a_30] \\& =15(a_1+a_1+29 d)=15(2 a_1+29 d) \\& Also, T=_i=1^15 a_(2 i-1)=a_1+a_3++a_29 \\& =a_1+(a_1+2 d)++(a_1+28 d) \\& =15 a_1+2 d(1+2++14)=15 a_1+210 daligned , S-2 T=75aligned& 15(2 a_1+29 d)-2(15 a_1+210 d)=75 d=5 \\& a_5=27 a_1+4 d=27 a_1=7aligned , a_10=a_1+9 d=7

Let a_{1},a_{2},,a_{30} be an A.P., S = \sum_{i = 1}^{30}\mspace{2mu} a_{i} and T = \sum_{i = 1}^{15}\mspace{2mu} a_{(2i - 1)}. If a_{5} = 27 and S - 2T = 75, then a_{10} is equal to: Sequence & Series (Mathematics)

(b) : Let

d

is the common difference of the A.P.\n

\n\begin{aligned}\n& \therefore S=\sum_{i=1}^{30} a_i=\frac{30}{2}\left[a_1+a_{30}\right] \\\n& =15\left(a_1+a_1+29 d\right)=15\left(2 a_1+29 d\right) \\\n& \text { Also, } T=\sum_{i=1}^{15} a_{(2 i-1)}=a_1+a_3+\cdots+a_{29} \\\n& =a_1+\left(a_1+2 d\right)+\cdots+\left(a_1+28 d\right) \\\n& =15 a_1+2 d(1+2+\cdots+14)=15 a_1+210 d\n\end{aligned}\n

\n \nNow,

S-2 T=75

\n\begin{aligned}\n& \Rightarrow 15\left(2 a_1+29 d\right)-2\left(15 a_1+210 d\right)=75 \Rightarrow d=5 \\\n& a_5=27 \Rightarrow a_1+4 d=27 \Rightarrow a_1=7\n\end{aligned}\n

\n \nNow,

a_{10}=a_1+9 d=7+9(5)=52

Explanation

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