JEE Main 2024 — Sequences And Series Question with Solution

To solve this problem, we will start by using the properties of an arithmetic progression (AP).

The sum of the first $n$ terms of an AP can be calculated using the formula: $$S_n=\frac{n}{2}(2a+(n-1)d)$$ where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $d$ is the common difference between the terms.

Given the information: $$S_{10}=390$$

We can plug $n=10$ into the sum formula to get:

$$S_{10}=\frac{10}{2}(2a+(10-1)d)$$

$$390=5(2a+9d)$$

$$390=10a+45d$$

$$78=2a+9d.........\text{(1)}$$

Next, we're given the ratio of the tenth term ($T_{10}$) to the fifth term ($T_5$): $$\frac{T_{10}}{T_5}=\frac{15}{7}$$

The $n$th term of an AP is given by:

$$T_n=a+(n-1)d$$

So, for the tenth term: $$T_{10}=a+(10-1)d=a+9d$$

And for the fifth term:

$$T_5=a+(5-1)d=a+4d$$

Now we can write the ratio as:

$$\frac{a+9d}{a+4d}=\frac{15}{7}$$

$$7(a+9d)=15(a+4d)$$

$$7a+63d=15a+60d$$

$$63d-60d=15a-7a$$

$$3d=8a.........\text{(2)}$$

Now we have two equations (1) and (2):

$$78=2a+9d\text{(1)}$$

$$3d=8a\text{(2)}$$

We can solve these equations simultaneously.

From equation (2):

$$d=\frac{8}{3}a$$

Plugging this back into (1):

$$78=2a+9\left(\frac{8}{3}a\right)$$

$$78=2a+24a$$

$$78=26a$$

$$a=3$$

Now we can find $d$:

$$d=\frac{8}{3}a$$

$$d=\frac{8}{3}\times 3$$

$$d=8$$

Now we can find $S_{15}$ and $S_5$ using the formula for the sum of an AP.

For $S_{15}$:

$$S_{15}=\frac{15}{2}(2\cdot 3+(15-1)\cdot 8)$$

$$S_{15}=\frac{15}{2}(6+14\cdot 8)$$

$$S_{15}=\frac{15}{2}(6+112)$$

$$S_{15}=\frac{15}{2}\cdot 118$$

$$S_{15}=15\cdot 59$$

$$S_{15}=885$$

For $S_5$:

$$S_5=\frac{5}{2}(2\cdot 3+(5-1)\cdot 8)$$

$$S_5=\frac{5}{2}(6+4\cdot 8)$$

$$S_5=\frac{5}{2}(6+32)$$

$$S_5=\frac{5}{2}\cdot 38$$

$$S_5=5\cdot 19$$

$$S_5=95$$

The difference $S_{15}-S_5$ is:

$$S_{15}-S_5=885-95$$

$$S_{15}-S_5=790$$

Therefore, the correct answer is Option C, which is 790.