Compound Interest (CI) is the interest earned on the principal amount and also on the accumulated interest of the previous periods.

Formula for Compound Interest (CI) = P (1 + r/n) ^ nt
Where, P = Principal Amount
r = Rate of Interest
n = Number of times the interest is compounded
t = Time in years

Given,
P = Rs 6950
r1 = 6% p.a. for the first two years
r2 = 9% p.a. for the third year
n = 2 (half-yearly)
t = 3 years

Calculation:
Compound Interest (CI) for first two years
= P (1 + r1/n) ^ nt
= Rs 6950 x (1+ 6/2) ^ (2 x 3)
= Rs 6950 x (1 + 3) ^ 6
= Rs 6950 x 729
= Rs 50,530

Compound Interest (CI) for third year
= P (1 + r2/n) ^ nt
= Rs 6950 x (1 + 9/2) ^ (2 x 1)
= Rs 6950 x (1 + 4.5) ^ 2
= Rs 6950 x 20.25
= Rs 140,612.50

Total Compound Interest (CI)
= CI for first two years + CI for third year
= Rs 50,530 + Rs 140,612.50
= Rs 191,142.50

Compound Interest (CI) for 3 years
= Total CI - Principal Amount
= Rs 191,142.50 - Rs 6950
= Rs 184,192.50

Compound Interest (CI) for 3 years
= CI for 3 years - CI for first two years
= Rs 184,192.50 - Rs 50,530
= Rs 133,662.50

Therefore, the compound interest on Rs 6950 for 3 years if interest is payable half-yearly, at the rate of 6% p.a. for the first two years and at the rate of 9% p.a. for the third year is Rs 1590.

If you think the solution is wrong then please provide your own solution below in the comments section .

Amount = Rs.\(\left[1600\times\left(1+\frac{5}{2\times100}\right)^{2}+1600\times\left(1+\frac{5}{2\times100}\right)\right]\)

= Rs. \(\left[1600\times\frac{41}{40}\times\frac{41}{40}+1600\times\frac{41}{40}\right]\)

= Rs. \(\left[1600\times\frac{41}{40}\left(\frac{41}{40}+1\right)\right]\)

= Rs. \(\left[\frac{1600\times41\times81}{40\times40}\right]\)

= Rs. 33.21

So, C.I. = Rs. (3321 - 3200) = Rs. 121

Let the sum be Rs. x. Then, 

C.I. =  \(\left[x\left(1\frac{4}{100}\right)^{2}-x\right] = \left(\frac{676}{925}x-x\right) = \frac{51}{625}x.\)

S.I. =  \(\left(\frac{x\times4\times2}{100}\right)= \frac{2x}{25.}\)

So, \(\frac{51x}{625}-\frac{2x}{25}=1\)

x=625.

Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.

So, R =    \(\left(\frac{100\times60}{100\times6}\right)\)   =10% p.a.

Now, P = Rs. 12000. T = 3 years and R = 10% p.a.

So, C.I. = Rs.  \(\left[12000\times\left\{\left(1+\frac{10}{100}\right)^{3}-1\right\}\right]\)

= Rs. \(\left(12000\times\frac{331}{1000}\right)\)

= 3972

C.I. when interest compounded yearly  = Rs. \(\left[5000\times\left(1+\frac{4}{100}\right)\times\left(1+\frac{\frac{1}{2}\times4}{100}\right)\right]\)

= Rs. \(\left(5000\times\frac{26}{25}\times\frac{51}{50}\right)\)

= Rs 5304.

C.I. when interest compounded half- yearly = Rs. \(\left[5000\times\left(1+\frac{2}{100}\right)^{3}\right]\)

= Rs \(\left(5000\times\frac{51}{50}\times\frac{51}{50}\times\frac{51}{50}\right)\)

= Rs. 5306.4

 So, Difference = Rs. (5306.04 - 5304) = Rs. 2.04

Amount = Rs. (30000 + 4347) = Rs. 34347.

Let the time be n years.

Then ,  \(30000\left(1+\frac{7}{100}\right)^{n}= 34347\)

\(\Rightarrow\left(\frac{107}{100}\right)^{n}=\frac{34347}{30000}=\frac{11449}{10000}=\left(\frac{107}{100}\right)^{2}\)

So, n = 2 years.