Answer: Option D. -> 2(n+1)n+2
:
D
$T_n=\frac{1}{\left[\frac{n(n+1)}{2}\right]}=2[\frac{1}{n}-\frac{1}{n+1}]$
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}$

Answer: Option B. -> 37
:
B
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. = $\frac{3}{7}$.

Answer: Option C. -> bc = ad
:
C
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.

Answer: Option D. -> (SR)n
:
D
$S=a\frac{(1-r^n)}{1-r}P=a^{n}r\frac{n(n-1)}{2}R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}+\cdots \cdots +\frac{1}{a{r}^{n-1}}=\frac{\frac{1}{a}(1-\frac{1}{r^n})}{1-\frac{1}{r}}=\frac{1}{a{r}^{n-1}}\left(\frac{r^n-1}{r-1}\right)p^2=a^{2n}{r}^{n(n-1)}\left(\frac{S}{R}\right)=a^{2n}{r}^{n(n-1)}$

Answer: Option A. -> 7
:
A
(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.........

Answer: Option A. ->   2516 -   4n+516×5n−1
:
A
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 - $\frac{n}{5^n}$
⇒ $\frac{4}{5}$ $S_n$ = $\frac{1-\frac{1}{5^n}}{\frac{4}{5}}$ - $\frac{n}{5^n}$⇒ $S_n$ = $\frac{25}{16}$ - $\frac{4n+5}{16\times 5^{n-1}}$

Answer: Option C. -> 4
:
C
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

Answer: Option A. -> 2
:
A
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

Answer: Option C. -> Rs. 20500
:
C
It will take 10 years for harikiranto 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.

Answer: Option D. -> 5615
:
D
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 $\frac{56}{15}$