If a_1,a_2,a_3 ... and b_1,b_2,b_3 ... are in A.P., and a_1 = 2, a_10 = 3,a_1b_1 = 1 = a_10b_10, then a_4b_4 is equal to

(d): We have given, a_1=2, a_10=3, a_1 b_1=1=a_10 b_10Let d_1 be the common difference of an A.P. a_1, a_2, a_3, ... ... and d_2 be the common difference of an A.P. b_1, b_2, b_3 ... ... a_10=a_1+9 d_1 => 3=2+9 d_1 => d_1=(1 / 9) \ b_10=b_1+9 d_2 => (1 / 3)=(1 / 2)+9 d_2 => d_2=-(1 / 54) \ a_4=a_1+3 d_1=2+3((1 / 9))=(7 / 3) \ b_4=b_1+3 d_2=(1 / 2)+3((-1 / 54))=(8 / 18) Therefore a_4 b_4=(28 / 27)\

If a_{1},a_{2},a_{3} and b_{1},b_{2},b_{3} are in A.P... | Sequence & Series (Mathematics)

(d): We have given, $a_1=2, a_{10}=3, a_1 b_1=1=a_{10} b_{10}$ Let $d_1$ be the common difference of an A.P. $a_1, a_2, a_3, \ldots \ldots$ and $d_2$ be the common difference of an A.P. $b_1, b_2, b_3 \ldots \ldots$ $\begin{aligned} & a_{10}=a_1+9 d_1 \Rightarrow 3=2+9 d_1 \Rightarrow d_1=\frac{1}{9} \\ & b_{10}=b_1+9 d_2 \Rightarrow \frac{1}{3}=\frac{1}{2}+9 d_2 \Rightarrow d_2=-\frac{1}{54} \\ & a_4=a_1+3 d_1=2+3\left(\frac{1}{9}\right)=\frac{7}{3} \\ & b_4=b_1+3 d_2=\frac{1}{2}+3\left(\frac{-1}{54}\right)=\frac{8}{18} \therefore a_4 b_4=\frac{28}{27}\end{aligned}$

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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