Free Objective Test 01 Practice Test - 11th and 12th 

Obviously, $7^{th}$ term of corresponding A.P is $\frac{1}{8}$ and $8^{th}$

term will be $\frac{1}{7}$. a + 6d = $\frac{1}{8}$ and a+7d = $\frac{1}{7}$

Solving these, we get d = $\frac{1}{56}$ and a = $\frac{1}{56}$

Therefore ${15}^{th}$ term of this A.P.

= $\frac{1}{56}+14\times \frac{1}{56}$ = $\frac{15}{56}$

Hence the required ${15}^{th}$ term of the H.P. is $\frac{56}{15}$

Let there be 2n terms in the given G.P. with first term a and the common ratio r.

Then, a  $\frac{r^2-1}{(r-1)}$ = 5a $\frac{(r^2-1)}{(r^2-1)}$ ⇒ r + 1 = 5 ⇒ r = 4

$S=a\frac{(1-r^n)}{1-r}P=a^{n}r\frac{n(n-1)}{2}R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\cdots \cdots +\frac{1}{a{r}^{n-1}}=\frac{\frac{1}{a}(1-\frac{1}{r^n})}{1-\frac{1}{r}}=\frac{1}{a{r}^{n-1}}\left(\frac{r^n-1}{r-1}\right)p^2=a^{2n}{r}^{n(n-1)}\left(\frac{S}{R}\right)=a^{2n}{r}^{n(n-1)}$

(b, c, d) Given $x_n$ = $x_{n+1}$ $\sqrt{2}$

∴ $x_1$ = $x_2$$\sqrt{2}$, $x_2$ = $x_3$$\sqrt{2}$, $x_n$ = $x_{n+1}$$\sqrt{2}$

On multiplying $x_1$ = $x_{n+1}$ ${\left(\sqrt{2}\right)}^n$   ⇒ $x_{n+1}$ =   $\frac{x_1}{(\sqrt{2}{)}^n}$

Hence  $x_n$ =   $\frac{x_1}{(\sqrt{2}{)}^{n-1}}$

Area of $S_n$ = $x_n^2$ =  $\frac{x_n^2}{2^{n-1}}$ < 1  ⇒  $2^{n-1}$ > $x_1^2$   ($x_1$ = 10)

∴ $2^{n-1}$ > 100

But $2^7$ > 100, $2^8$>100, etc.

∴ n - 1 = 7, 8, 9.......  ⇒ n = 8, 9, 10.........

It will take 10 years for harikiran to pay off Rs. 10000 in 10 yearly

installments.

∵ He pays 10% annual interest on interest on remaining amount

∴ Money given in first year

= 1000 +  $\frac{10000\times 10}{100}$ = Rs. 2000

Money given in second year = 1000 + interest of (10000 - 1000) with interest rate 10% per annum

= 1000 +  $\frac{9000\times 10}{100}$ = Rs. 1900

Money paid  in third year = Rs. 1800 etc.

So money given by Jairam in 10 years will be Rs. 2000, Rs.1900, Rs. 1800, Rs. 1700,......,

which is in arithmetic progression,

whose first term a = 2000 and d = -100

Total money given in 10 years = sum of 10 terms of arithmetic 

progression

=  $\frac{10}{2}$[2(2000) + (10 - 1)(-100)] = Rs. 15500

Therefore, total money given by harikiran 

= 5000 + 15500 = Rs. 20500.

$T_n=\frac{1}{\left[\frac{n(n+1)}{2}\right]}=2[\frac{1}{n}-\frac{1}{n+1}]$

  put n=1,2,3,.........,(n+1)

   $T_1=2[\frac{1}{1}-\frac{1}{2}]$,$T_2=2[\frac{1}{2}-\frac{1}{3}]$,...........,

   $T_{n+1}=2[\frac{1}{n+1}-\frac{1}{n+2}]$

Hence sum of (n+1) terms = $\sum _{k=1}^{n+1}{T}_k$ 

 $\Rightarrow$  $S_{n+1}=2[1-\frac{1}{n+2}]$ $\Rightarrow$   $S_{n+1}=\frac{2(n+1)}{n+2}$

$2^{\frac{1}{4}}$. $4^{\frac{1}{8}}$. $8^{\frac{1}{16}}$. ${16}^{\frac{1}{32}}$ .........∞

=  $2^{\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+........}$ =  $2^S$, where S is given by 

S =  $\frac{1}{4}$ + 2 $\frac{1}{8}$ + 3 $\frac{1}{16}$ +4 $\frac{1}{32}$ + ............∞                              .........(i)

⇒  $\frac{1}{2}$ S =  $\frac{1}{8}$ +  $\frac{2}{16}$ + $\frac{3}{32}$ + $\frac{4}{64}$ + ................∞     ..........(ii)

Subtracting (ii) from (i), we get S = 1.

Hence required product =  $2^1$ = 2.

Given series, let $S_n$ = 1 +  $\frac{2}{5}$ +  $\frac{3}{5^2}$ +  $\frac{4}{5^3}$ + .............+  $\frac{n}{5^{n-1}}$

                        $\frac{1}{5}$$S_n$ =  $\frac{1}{5}$ +  $\frac{2}{5^2}$ +  $\frac{3}{5^3}$ + ..............+  $\frac{n}{5^n}$

Subtracting,

(1 - $\frac{1}{5}$)$S_n$ = 1 +  $\frac{1}{5}$ +  $\frac{1}{5^2}$ +  $\frac{1}{5^3}$ + ............+ upto n terms -  $\frac{n}{5^n}$

⇒  $\frac{4}{5}$ $S_n$ =  $\frac{1-\frac{1}{5^n}}{\frac{4}{5}}$ -  $\frac{n}{5^n}$ ⇒ $S_n$ =   $\frac{25}{16}$ -   $\frac{4n+5}{16\times 5^{n-1}}$

The sum of the first n terms is

$S_n$ = (1 - $\frac{1}{2}$) + (1 - $\frac{1}{2^2}$) + (1 - $\frac{1}{2^3}$) + (1 - $\frac{1}{2^4}$  + ........... + (1 - $\frac{1}{2^n}$)

= n - { $\frac{1}{2}$ +  $\frac{1}{2^2}$ +............+  $\frac{1}{2^n}$}

= n -  $\frac{1}{2}$($\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}$) = n - (1 - $\frac{1}{2^n}$) = n - 1 +  $2^{-n}$.

Trick: Check for n = 1, 2 i.e., $S_1$ =  $\frac{1}{2}$, $S_2$ =  $\frac{5}{4}$ and

(c)  ⇒ $S_1$ =   $\frac{1}{2}$ and $S_2$ = 2 +  $2^{-2}$  - 1 =  $\frac{5}{4}$.

Let $S_n$ be the sum of the given series to n terms, then 

$S_n$ = 1 + 2x + 3 $x^2$ + 4 $x^3$ + ....... + n $x^{n-1}$                         ..........(i)

x$S_n$ =       x + 2 $x^2$ + 3 $x^2$ + ...........+ n $x^n$                          ..........(ii)

Subtracting (ii) from (i), we get 

(1-x)$S_n$ = 1 + x +  $x^2$ +  $x^3$ + ......... to n terms -n $x^n$

                   = ( $\frac{(1-x^n)}{(1-x)}$) - n$x^n$

⇒ $S_n$ =  $\frac{(1-x^n)-n{x}^n(1-x)}{(1-x{)}^2}$ =  $\frac{1-(n+1)x^n+n{x}^{n+1}}{(1-x{)}^2}$

Given that a,b,c in A.P. and b,c,d in H.P.

So, 2b = a + c and c = $\frac{2bd}{b+d}$

$\Rightarrow$ c(b + d) = 2bd = (a + c)d $\Rightarrow$ bc = ad.

Let A = $\frac{a+b}{2}$, G =$\sqrt{ab}$ and H = $\frac{2ab}{a+b}$.

Then $G^2$ = ab            ..................(i)

and AH = $\left(\frac{a+b}{2}\right).\frac{2ab}{a+b}$ = ab    ................(ii)

From (i) and (ii), we have $G^2$ = AH

As given H = $\frac{2pq}{p+q}$

$\therefore \frac{H}{p}+\frac{H}{q}$ = $\frac{2q}{p+q}+\frac{2p}{p+q}$ = $\frac{2(p+q)}{p+q}$ = 2

a + 3d = $\frac{5}{3}$ and a + 7d = 3

Solving a = $\frac{2}{3}$, d = $\frac{1}{3}$

6th term of A.P = a + 5d = $\frac{2}{3}+\frac{5}{3}$ = $\frac{7}{3}$

$\Rightarrow$ 6th term of H.P. = $\frac{3}{7}$.

Given that
$a_{p+q}=a{r}^{p+q-1}=m$ and $a_{p-q}=a{r}^{p-q-1}=n.$

$\Rightarrow m\times n=a{r}^{p+q-1}$ x  $a{r}^{p-q-1}$

                 $=a^2{r}^{2(p-1)}$ 

                 $=(a{r}^{p-1}{)}^2$

 $\Rightarrow \sqrt{mn}=a{r}^{p-1}=a_p$   

 Thus, the $p^{\text{th}}$ term of the GP is $\sqrt{mn}.$

Aliter: Each term in a G.P. is the geometric mean of the terms equidistant from it. Here, $(p+q{)}^{th}$ and $(p-q{)}^{th}$ terms are equidistant from the $p^{th}$ term. 
$\therefore$ $p^{th}$ term ($\sqrt{mn}$) will be G.M. of $(p+q{)}^{th}$ and $(p-q{)}^{th}$ terms.