Let a_1,a_2, ... ,a_30 be an A.P.,n S = sum _i = 1^30mspace2mu a_i andn T = sum _i = 1^15mspace2mu a_(2i - 1). If a_5 = 27n and S - 2T = 75, then a_10 is equal to

(b) : Let d is the common difference of the A.P.nn n Therefore S=sum _i=1^30 a_i=(30 / 2)[a_1+a_30] n =15(a_1+a_1+29 d)=15(2 a_1+29 d) n Also, T=sum _i=1^15 a_(2 i-1)=a_1+a_3+ * s+a_29 n =a_1+(a_1+2 d)+ * s+(a_1+28 d) n =15 a_1+2 d(1+2+ * s+14)=15 a_1+210 dn nn nNow, S-2 T=75nn n => 15(2 a_1+29 d)-2(15 a_1+210 d)=75 => d=5 n a_5=27 => a_1+4 d=27 => a_1=7n nn nNow, a_10=a_1+9 d=7+9(5)=52

Let a_{1},a_{2},,a_{30} be an A.P., S = \sum_{i = 1}^... | 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

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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