Question

Find the sum of indicated number of terms in each of the following APs
() 16,11,6....., 23 terms (i) -0.5, -1.0,-1.5.......10 terms
_1,1/4,3/2..…....10 terms​

Answer 1

Answer:

1 formula n/2 (2a +(n-1)d) se, a =16, d=5, n= 23 rakhane par answer 1633 ayega

2 a= -0.5, d=0.5 n=10 rakne par answer 17.5 answer ayega

Answer 2

(i)

Given:

• A.P = 16,11,6, ...
• a = 16,
• d = $a_{2} - a_{1} = 11 - 16 = -5$
• n = 23

We know that,

$\qquad \quad \; \boxed{S_{n} = \frac{n}{2}[ \: 2a + (n-1)d \:} \\ \Rightarrow \qquad \frac{23}{2}[\: 2 \cdot 16 + (23-1)-5 \: ] \\ \Rightarrow \qquad \frac{23}{2}( 32 + 22 \cdot -5) \\ \Rightarrow \qquad \frac{23}{2} \times -78 \\ \qquad \quad \boxed{S_{23} = -897}$

Answer: -897

(ii)

Given:

• A.P = -0.5, -1.0, -1.5 , ...
• a = -0.5
• d = $a_{2} - a_{1} = -1.0 + 0.5 = -0.5$
• [n = 10

We know that,

$\qquad \quad \; \boxed{S_{n} = \frac{n}{2}[ \: 2a + (n-1)d \:} \\ \Rightarrow \qquad \frac{10}{2}[ \: 2 \cdot -0.5 + (10-1)-0.5 \: ] \\ \Rightarrow \qquad 5(-1.0 + 9 \times -0.5) \\ \Rightarrow \qquad 5 \times -5.5 \\ \qquad \quad \; \boxed{S_{10} = -27.5}$

Answer: -27.5

(iii)

Given:

• A.P = -1, 1/4, 3/2 , ...
• a = -1
• d = $a_{2} - a_{1} = \frac{1}{4} -(-1) \Rightarrow \frac{1+4}{4} = \frac{5}{4}$
• n = 10

We know that,

$\qquad \quad \; \boxed{S_{n} = \frac{n}{2}[ \: 2a + (n-1)d \:} \\ \Rightarrow \qquad \frac{10}{2}\left( \: 2 \cdot -1 + (10-1) \times \frac{5}{4} \right) \\ \Rightarrow \qquad 5\left( -2 + 9 \times \frac{5}{4} \right) \\ \Rightarrow \qquad 5 \times \frac{37}{4} \\ \qquad \quad \; \boxed{ S_{10} = \frac{185}{4}}$

Answer : $\bold{\left( \frac{185}{4} \right)}$