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Solve the Following Question.(3 Marks)

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19 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Find $\left(50^2-49^2\right)+\left(48^2-47^2\right)+\left(46^2-45^2\right)+\ldots \ldots+\left(2^2-1^2\right)$
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Question 33 Marks
For a sequence, $S_n=4\left(7^n-1\right)$, verify whether the sequence is a G.P.
Answer
$
\begin{aligned}
& \mathrm{S}_{\mathrm{n}}=4\left(7^{\mathrm{n}}-1\right) \\
& \therefore \quad \mathrm{S}_{\mathrm{n}-1}=4\left(7^{\mathrm{n}-1}-1\right) \\
& \text { But, } \mathrm{t}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1} \\
& =4\left(7^n-1\right)-4\left(7^{n-1}-1\right) \\
& =4\left(7^{\mathrm{n}}-1-7^{\mathrm{n}-1}+1\right) \\
& =4\left(7^{\mathrm{n}}-7^{\mathrm{n}-1}\right) \\
& =4\left(7^{\mathrm{n}-1+1}-7^{\mathrm{n}-1}\right) \\
& =4.7^{\mathrm{n}-1}(7-1) \\
& \therefore \quad \mathrm{t}_{\mathrm{n}}=24.7^{\mathrm{n}-1} \\
& \therefore \quad \mathrm{t}_{\mathrm{n}-1}=24.7^{(\mathrm{n}-1)-1}=24.7^{\mathrm{n}-2} \\
&
\end{aligned}
$
The sequence is a G.P., if $\frac{t_n}{t_{n-1}}=$ constant for all $\mathrm{n} \in \mathrm{N}$.
$
\begin{aligned}
\therefore \quad \frac{\mathrm{t}_{\mathrm{n}}}{\mathrm{t}_{\mathrm{n}-1}} & =\frac{24.7^{\mathrm{n}-1}}{24.7^{\mathrm{n}-2}}=\frac{7^{\mathrm{n}-1}}{7^{\mathrm{n}-1} \cdot 7^{(-1)}} \\
& =7=\text { constant, for all } \mathrm{n} \in \mathrm{N}
\end{aligned}
$
$\therefore \quad$ the sequence is a G.P.
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Question 53 Marks
Find $\left(70^2-69^2\right)+\left(68^2-67^2\right)+\left(66^2-65^2\right)+\ldots \ldots . .+\left(2^2-1^2\right)$
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Question 93 Marks
For a sequence, if $S_n=2\left(3^{n-1}\right)$, find the $n$th term, hence showing that the sequence is a G.P.
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Question 183 Marks
If p, q, r, s are in G. P., show that p + q, q + r, r + s are also in G.P.
Answer
$p, q, r, s$ are in G.P.
$
\begin{array}{ll}
\therefore \quad & \frac{\mathrm{q}}{\mathrm{p}}=\frac{\mathrm{r}}{\mathrm{q}}=\frac{\mathrm{s}}{\mathrm{r}} \\
& \text { Let } \frac{\mathrm{q}}{\mathrm{p}}=\frac{\mathrm{r}}{\mathrm{q}}=\frac{\mathrm{s}}{\mathrm{r}}=\mathrm{k} \\
\therefore & \mathrm{q}=\mathrm{pk}, \mathrm{r}=\mathrm{qk}, \mathrm{s}=\mathrm{rk}
\end{array}
$
We have to prove that $\mathrm{p}+\mathrm{q}, \mathrm{q}+\mathrm{r}, \mathrm{r}+\mathrm{s}$ are in G.P. i.e., to prove that $\frac{q+r}{p+q}=\frac{r+s}{q+r}$
$
\begin{aligned}
& \text { L.H.S. }=\frac{q+r}{p+q}=\frac{q+q k}{p+p k}=\frac{q(1+k)}{p(1+k)}=\frac{q}{p}=k \\
& \text { R.H.S. }=\frac{r+s}{q+r}=\frac{r+r k}{q+q k}=\frac{r(1+k)}{q(1+k)}=\frac{r}{q}=k \\
& \therefore \quad \frac{q+r}{p+q}=\frac{r+s}{q+r} \\
& \therefore p+q, q+r, r+s \text { are in G.P. }
\end{aligned}
$
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Question 193 Marks
Find five numbers in G. P. such that their product is 1024 and the fifth term is square of the third term.
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