Question

in an arithmetic progression of 15 terms the sum of first 10 terms is 210 and the sum of last 15 term 2565 find the automatic progression

Answer 1

Dear Student,

Assuming your question as :

In an arithmetic progression of 50 terms, the sum of first 10 terms is 210 and the sum of last 15 term 2565 find the arithmetic progression.

Here is the answer.

Let a be the first term and d be the common difference of the given A.P
We know that sum of first n terms of an A.P is given by,
$$\begin{gathered} {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right] \\ \mathrm{Put}\mathrm{n}=10,\mathrm{we}\mathrm{get} \\ {\mathrm{S}}_{10}=\frac{10}{2}\left[2\mathrm{a}+9\mathrm{d}\right] \\ \Rightarrow 210=5\left(2\mathrm{a}+9\mathrm{d}\right) \\ \Rightarrow 2\mathrm{a}+9\mathrm{d}=42...........\left(1\right) \\ \mathrm{Sum}\mathrm{of}\mathrm{last}15\mathrm{terms}=2565 \\ \Rightarrow \mathrm{Sum}\mathrm{of}\mathrm{first}50\mathrm{terms}-\mathrm{sum}\mathrm{of}\mathrm{first}35\mathrm{terms}=2565 \\ \Rightarrow {\mathrm{S}}_{50}-{\mathrm{S}}_{35}=2565 \\ \Rightarrow \frac{50}{2}\left[2\mathrm{a}+(50-1)\mathrm{d}\right]-\frac{35}{2}\left[2\mathrm{a}+(35-1)\mathrm{d}\right]=2565 \\ \Rightarrow 25(2\mathrm{a}+49\mathrm{d})-\frac{35}{2}\left(2\mathrm{a}+34\mathrm{d}\right)=2565 \\ \Rightarrow 5(2\mathrm{a}+49\mathrm{d})-\frac{7}{2}\left(2\mathrm{a}+34\mathrm{d}\right)=513 \\ \Rightarrow 3\mathrm{a}+126\mathrm{d}=513 \\ \Rightarrow \mathrm{a}+42\mathrm{d}=171...........\left(2\right) \\ \mathrm{multiply}\left(2\right)\mathrm{with}\text{'}2\text{'}\mathrm{we}\mathrm{get} \\ 2\mathrm{a}+84\mathrm{d}=342..........\left(3\right) \\ \mathrm{subtracting}\left(1\right)\mathrm{from}\left(3\right),\mathrm{we}\mathrm{get} \\ 84\mathrm{d}-9\mathrm{d}=342-42 \\ \Rightarrow 75\mathrm{d}=300 \\ \Rightarrow \mathrm{d}=\frac{300}{75}=4 \\ \mathrm{now},\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get} \\ \mathrm{a}+42\left(4\right)=171 \\ \mathrm{a}+168=171 \\ \mathrm{a}=3 \end{gathered}$$

Now, a₅₀ = a + 49d = 3 + 49 × 4 = 3 + 196 = 199
so, the AP is a, a+d, a+2d, a+3d, ........., a+49d

so the required AP is, 3, 7, 11, 15, ......., 199

Regards