Question

Q1.If the sum of first four terms of an AP is 40 and that of first 14 terms is 280.
Find the sum of its firstn terms
36.​

Answer 1

Explanation:

Given

We have given S₄= 40 and S₁₄= 280

To find

We have to find the sum of its nth term

SOLUTION:

We will apply sum formula

sₙ= n/2 [2a +(n-1)d]

⇢s₄= 4/2(2a+(4-1)d

⇢s₄= 2(2a+3d)

⇢40= 2(2a+3d)

20= 2a+3d ----(1)

Now,

⇢S₁₄= 14/2(2a+(14-1)d

⇢280= 7(2a+13d)

40= 2a+13d-----(2)

Now, Substract Equation 1 from Equation 2

=>40-20= 2a+13d-(2a+3d)

=>20= 2a+13d-2a-3d

=>20= 10d

=>d= 20/10=2

hence, common difference (d) is 2

substitute this value into equation 1

=>20= 2a+3(2)

=>20= 2a+6

=>2a= 14

=>a= 14/2=7

a= 7

Now,we have to find sum of first n terms

⇢sₙ=n/2(2a+(n-1)d

⇢sₙ= n/2(2(7)+(n-1)2

⇢sₙ= n/2(14+2n-2)

⇢sₙ=n/2(12+2n)

⇢sₙ= n/2 [ 2(6+n)]

⇢sₙ= n(n+6)

sₙ= n²+6n

Hence,sum of its n terms is n²+6n

Answer 2

Answer:

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Explanation:

$we \: know \: that \: {sn}^ = { \frac{n }{2} } (2a + (n - 1)d) \\ {s4 = 40 = \frac{4}{2 } }^{} (2a + (4 - 1)d) = 40 \\ = 2a + 3d = 20 - - (1) \\ and {3}^{14} = 280 = \frac{14}{2} (2a + (14 - 1)d) = 280 \\ = 2a + 13d = - - (2) \\ after \: solving \: (1)and \: (2)we \: get \\ a = 7and \: d = 2 \\ {s}^{n = \frac{n}{2 } } (2 \times 7 + (n - 1)2) \\ = {s}^{n } = (7 + n - 1) = n(n + 6) \\ = {s}^{n } = n(n + 6)$