Question

a13 is 4 times a3 , if a5=16 find s10

Answer 1

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Answer 2

S₁₀ = 175

Given :

a₁₃ is 4 times a₃

a₅ = 16

To find :

The value of S₁₀

Formula Used :

1. The nth term of an AP is

aₙ = a + (n - 1 )d.

a = first term

aₙ = nth term

d = common difference.

2. Sum of first n terms of an arithmetic progression

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

Where First term = a

Common Difference = d

Solution :

Step 1 of 2 :

Form the equation to find first term and common difference

Let first term = a and common difference = d

a₁₃ = a + 12d

a₃ = a + 2d

a₅ = a + 4d

Now , a₁₃ is 4 times a₃

∴ a + 12d = 4(a + 2d)

⇒ a + 12d = 4a + 8d

⇒ 4d = 3a - - - - - (1)

a₅ = 16 gives

a + 4d = 16 - - - - - (2)

Step 2 of 2 :

Calculate first term and common difference

From Equation 1 and Equation 2 we get

a + 3a = 16

⇒ 4a = 16

⇒ a = 4

From Equation 2 we get

4 + 4d = 16

⇒ 4d = 12

⇒ d = 3

Step 3 of 3 :

Calculate the value of S₁₀

S₁₀

= Sum of first 10 terms

$\displaystyle \sf = \frac{10}{2} \bigg[2a + (10 - 1)d\bigg]$

$\displaystyle \sf = 5 \times \bigg[(2 \times 4) + (9 \times 3)\bigg]$

$\displaystyle \sf = 5 \times \bigg[8 + 27\bigg]$

$\displaystyle \sf = 5 \times 35$

$\displaystyle \sf = 175$

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