Question

solve question no. 14

Answer 1

ANSWER:

Given:

• $a_{n} = 1 - 3n$

Since we know the nth-term, we can put n = 1 for first term , n = 2 for second term and so on.

$n = 1,2,3,....$

$a_{1} = 1 - 3 \cdot 1 = \bold{-2} \\ a_{2} = 1 - 3 \cdot 2 = \bold{-5} \\ a_{3} = 1 - 3 \cdot 3 = \bold{-8}$

Now , we know the first, second and third terms, We can find out the common difference, which is:

$\qquad \quad \; d = a_{2} - a_{1} = a_{3} - a_{2} \\ \Rightarrow \qquad d = -5 -(-2) = -8 -(-5) \\ \Rightarrow \qquad d = -5 + 2 = -8 + 5 \\ \Rightarrow \qquad d = \bold{-3}$

Now, We are asked to find the sum of first 25 terms:

$\qquad \quad \; \boxed{\because \quad S_{n} = \frac{n}{2}[2a + (n-1)d]} \\ \Rightarrow \qquad S_{25} = \frac{25}{2}[2 \cdot -2 + (25 - 1) \cdot -3 ] \\ \Rightarrow \qquad S_{25} = \frac{25}{2}(-4 + 24 \cdot -3) \\ \Rightarrow \qquad S_{25} = \frac{25}{2}(-4 - 72) \\ \Rightarrow \qquad S_{25} = \frac{25}{2} \times -76 \\ \Rightarrow \qquad S_{25} = -950$

$\huge{\underline{ANSWER} : -950}$