Question

please answer fast, class 10 exercise 5.3 question no. 3 (i) ​

Answer 1

Answer:

This is your answer :-

Step-by-step explanation:

n=16,

Sn=440

Answer 2

Given:-

• a = 5
• d = 3
• $\sf{a_n = 50}$

To find:-

n and $\sf{S_n}$

Solution:-

We know,

In an A.P.

$\sf{a_n = a + (n-1)d}$

Therefore,

= $\sf{50 = 5 + (n - 1)\times3}$

= $\sf{50 = 5 + 3n - 3}$

= $\sf{50 = 2+3n}$

= $\sf{50-2 = 3n}$

= $\sf{48 = 3n}$

= $\sf{n = \dfrac{48}{3}}$

= $\sf{n = 16}$

Therefore,

The value of n is 48

Now,

$\sf{S_n = \dfrac{n}{2}[2a + (n-1)d]}$

= $\sf{S_{16} = \dfrac{16}{2}[2\times5 + (16-1)\times 3]}$

= $\sf{S_{16} = 8[10+15\times3]}$

= $\sf{S_{16} = 8[10+45]}$

= $\sf{S_{16} = 8\times55}$

= $\sf{S_{16} = 440}$

Therefore,

$\sf{S_n = 440}$

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How did we get our answer?

For the first part.

We know the nth term of an A.P. is denoted by the formula:-

$\sf{a_n = a+(n-1)d}$

Where,

n = number of terms

a = First term of the A.P.

d = Common Difference

Here in this question we were given the value of a, d and $\sf{a_n}$. We were asked to find the value of n. So substituting all the values in the formula $\sf{a_n = a+(n-1)d}$.

For the second part.

We know,

Sum of nth term of an A.P. is denoted by the formula:-

$\sf{S_n = \dfrac{n}{2}[2a + (n-1)d]}$

Where the values of a, d and n are same.

We had the values a, n and d which we substituted in the formula and got the value of $\sf{S_n}$

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