4 Marks Questions

Question 14 Marks

If $a$ is A.M. of $b$ and $c$ and the two geometric means are $G_1$ and $G_2$, then prove that $G_1^3+G_2^3=2 a b c $

Answer

It is given that $a$ is the A.M. of $b$ and $c$.
$\therefore \quad a=\frac{b+c}{2} \Rightarrow b+c=2 a$
Since $G_1$ and $G_2$ are two geometric means between $b$ and $c$. Therefore, $b, G_1, G_2, c$ is a G.P. with common ratio $r=\left(\frac{c}{b}\right)^{\frac{1}{3}}$
$\therefore \quad G_1=b r=b\left(\frac{c}{b}\right)^{\frac{1}{3}}=c^{\frac{1}{3}} b^{\frac{2}{3}}$
and $G_2=b r^2=b\left(\frac{c}{b}\right)^{\frac{2}{3}}=b^{\frac{1}{3}} c^{\frac{2}{3}}$
$\begin{array}{rlrl}\Rightarrow\quad\quad G_1^3 & =b^2 c \text { and } G_2^3=b c^2 \\ \Rightarrow G_1^3+G_2^3 & =b^2 c+b c^2 \\ \Rightarrow G_1^3+G_2^3 & =b c(b+c) \\ \Rightarrow G_1^3+G_2^3 & =2 a b c\end{array}$
[using (i)]

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Question 24 Marks

The product of first three terms of G.P. is 1000 . If 6 is added to its second term and 7 is added to its third term, the terms become in A.P. Find G.P.

Answer

Let the numbers in G.P. be $\frac{a}{r}, a, a r \quad \ldots(i)$
$
\begin{aligned}
& & \text { Product } \frac{a}{r} \cdot a \cdot a r & =1000 \\
\Rightarrow & & a^3 & =1000 \\
\Rightarrow & & a & =10
\end{aligned}
$
According to question
$
\begin{array}{l}
a_1=\frac{a}{r}=\frac{10}{r} \\
a_2=a+6=10+6=16 \\
a_3=a r+7=10 r+7 \\
Also, a_3=a_1+2\left(a_2-a_1\right) \\
\quad\left[\because a_1, a_2, a_3 \text { are in A.P. }\right]
\end{array}
$
$\begin{array}{lrl}\Rightarrow 10 r+7 =\frac{10}{r}+2\left[16-\frac{10}{r}\right] \\ \Rightarrow 10 r^2+7 r =10+32 r-20 \\ \Rightarrow 10 r^2-25 r+10 =0 \\ \Rightarrow (r-2)(10 r-5)=0\end{array}$
$\Rightarrow \quad r=2 \text { or } \frac{1}{2}$
Substituting the value of $a$ and $r$ in eq (i), we get G.P. $: 5,10,20, \ldots$ when $r=2$
and G.P. : $20,10,5, \ldots$ when $r=\frac{1}{2}$

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Question 34 Marks

If $a, b, c$ are in A.P. show that $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right)$, $c\left(\frac{1}{a}+\frac{1}{b}\right)$ are also in A.P.

Answer

$a, b, c$ are in A.P.
$\begin{array}{l}\Rightarrow \frac{a}{a b c}, \frac{b}{a b c}, \frac{c}{a b c} \text { are in A.P. } \\ \text { [on dividing each term by } a b c \text { ] } \\ \Rightarrow \frac{1}{b c}, \frac{1}{c a}, \frac{1}{a b} \text { are in A.P. } \\ \Rightarrow \frac{a b+b c+c a}{b c}, \frac{a b+b c+c a}{c a}, \frac{a b+b c+c a}{a b} \text { are in A.P. }\end{array}$
[on multiplying each term by $a b+b c+c a$ ]
$\Rightarrow \frac{a b+b c+c a}{b c}-1, \frac{a b+b c+c a}{c a}-1, \frac{a b+b c+c a}{a b}-1$
are also in A.P.
[On adding - 1 to each term]
$\Rightarrow \frac{a b+a c}{b c}, \frac{a b+b c}{c a}, \frac{b c+c a}{a b}$ are in A.P.
$\Rightarrow a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P.

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Question 44 Marks

In an A.P., if the $p$ th term is $\frac{1}{q}$ and $q$ th term is $\frac{1}{p}$.
Prove that the sum of first $p q$ term is $\frac{1}{2}(p q+1)$.

Answer

$\because T_n=a+(n-1) d$
Therefore, $\quad T_p=a+(p-1) d=\frac{1}{q} \quad$ (given) . . . (i)
and $\quad a+(q-1)=\frac{1}{p}$ (given) . . . (ii)

Subtracting Eq. (i) from Eq.(ii),
$
\begin{aligned}
d(p-1-q+1) =\frac{1}{q}-\frac{1}{p} \\
\Rightarrow d(p-q) =\frac{p-q}{p q} \Rightarrow d=\frac{1}{p q}
\end{aligned}
$
Putting the value of $d$ in Eq. (i) we get
$
\begin{aligned}
& & a+\frac{(p-1)}{p q} & =\frac{1}{q} \\
\Rightarrow & & a & =\frac{1}{q}-\frac{p-1}{p q} \\
\Rightarrow & & a & =\frac{p-p+1}{p q}=\frac{1}{p q}
\end{aligned}
$
Now,
$
\begin{aligned}
\Rightarrow \quad S_{p q} & =\frac{p q}{2}[2 a+(p q-1) d] \\
& \left(\therefore S_n=\frac{n}{2}[2 a+(n-1) d]\right) \\
& =\frac{p q}{2}\left[2 \times \frac{1}{p q}+(p q-1) \frac{1}{p q}\right] \\
& =\frac{p q}{2} \times \frac{1}{p q}(2+p q-1) \\
& =\frac{1}{2}(p q+1)
\end{aligned}
$

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Question 54 Marks

The ratio of the sum of $n$ terms of two A.P.'s is $(7 n-1):(3 n+11)$. Find the ratio of their $10^{\text {th }}$ terms.

Answer

Let $a_1, a_2$ be the first terms and $d_1, d_2$ the common differences of the two given A.P's. Then, the sums of their $n$ terms are given by
$
\begin{array}{l}
S_n=\frac{n}{2}\left[2 a_1+(n-1) d_1\right] \text { and } S_n^{\prime}=\frac{n}{2}\left[2 a_2+(n-1) d_2\right] \\
\frac{S_n}{S_n^{\prime}}=\frac{\frac{n}{2}\left[2 a_1+(n-1) d_1\right]}{\frac{n}{2}\left[2 a_2+(n-1) d_2\right]}=\frac{2 a_1+(n-1) d_1}{2 a_2+(n-1) d_2}
\end{array}
$
It is given that,
$
\begin{aligned}
\frac{S_n}{S_n^{\prime}} & =\frac{7 n-1}{3 n+1} \\
\Rightarrow \quad \frac{2 a_1+(n-1) d_1}{2 a_2+(n-1) d_2} & =\frac{7 n-1}{3 n+11} \quad \ldots(i)
\end{aligned}
$
To find the ratio of the 10th terms of the two A.P.s', we replace $n$ by $(2 \times 10-1)$ i.e., '19' in (i),
$\therefore$ $\frac{2 a_1+(19-1) d_1}{2 a_2+(19-1) d_2}=\frac{7(19)-1}{3(19)+11}$
$\Rightarrow \quad \frac{a_1+9 d_1}{a_2+9 d_2}=\frac{133-1}{57+11}=\frac{132}{68}$
$\Rightarrow \quad \frac{a_1+9 d_1}{a_2+9 d_2}=\frac{33}{17}$
$\therefore$ Ratio of the $10^{\text {th }}$ terms of the two A.P.'s $=\frac{33}{17}$

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Question 64 Marks

If the $1^{\text {st }}, 2^{\text {nd }}$ and last terms of an A.P. are $a, b$ and $c$, respectively then find the sum of all terms of the A.P.

Answer

Given A.P. is $a, b, \ldots c$
First term $(A)=a$
Common difference $(d)=b-a$
Last term i.e., $n^{\text {th }}$ term $\left(A_n\right)=c$
Using formula, $\quad A_n=A+(n-1) d$, we get
$\Rightarrow \quad c=a+(n-1)(b-a)$
$\Rightarrow \quad c-a=(n-1)(b-a)$
$\Rightarrow \quad n-1=\frac{c-a}{b-a}$
$\Rightarrow \quad n=\frac{c-a}{b-a}+1$
$\Rightarrow \quad n=\frac{b+c-2 a}{b-a}\quad \ldots(i)$
Now, using formula $S_n=\frac{n}{2}\left(A+A_n\right)$ for sum of $n$ terms, we get$
\begin{aligned}
S_n & =\frac{n}{2}[a+c] \\
& =\frac{(b+c-2 a)(a+c)}{2(b-a)} \quad \text { using }(i) \\
& =\frac{(a+c)(b+c-2 a)}{2(b-a)}
\end{aligned}
$

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