Question

If the 5th and 12th terms of an AP −4 and −18 respectively, find the sum of first 20 terms of the AP.

Answer 1

Let a be the first term and d be the common difference of the given AP.
Then T₅ = -4 and T₁₂ = -18

⇒a + (5 - 1) d = -4 and a + (12 - 1) d = -18
i.e., a + 4d = - 4 ...(i)
also, a +11d = -18 ...(ii)
On subtracting (i) from (ii), we get:
7d = -14
⇒ d = -2
Putting d = -2 in (i), we get:
a = 4
∴ a = 4, d = -2 and n = 20
Sum of the first 20 terms = $\frac{n}{2}[2a+\left(n-1\right)d]$

$$\begin{gathered} =\left(\frac{20}{2}\right)\times \left[2\times 4+\left(20-1\right)\times \left(-2\right)\right] \\ =10\times \left[8-38\right]=10\times \left(-30\right)=-300 \end{gathered}$$

Hence, the required sum is -300.