Question

if the 5th term of an ap is -3 and its common difference is -4 . the sum of it'd first 10 terms is -
a)50
b)-50
c)30
d)-30

Answer 1

Given:-

• 5th term of an A.P is -3.

• Common difference is -4.

To Find:-

• Sum of it's first 10 terms

Formula used:-

• ${\boxed{a_n=a+(n-1)d}}$

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

Solution:-

Now,

$\implies\:a_n=a+(n-1)d$

Here,

• $a_5=-3$

• d = -4

• n = 5

$\implies\:a_5=a+(5-1)(-4)$

$\implies\:-3=a+(4)(-4)$

$\implies\:a=-3+16$

${\boxed{\implies\:a=13}}$

Now,

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

Here,

• a = 13

• n = 10

• d = -4

$\implies\:S_{10}=\dfrac{10}{2}[2 \times 13+(10-1)(-4)]$

$\implies\:S_{10}=5(26-36)$

$\implies\:S_{10}=5 \times (-10)$

${\boxed{\implies\:S_{10}=-50}}$

Hence, The Sum of first 10 Terms of the A.P is -50.

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Answer 2

GiveN :

5th term (a5) = -3

Common Difference (d) = -4

Number of terms (n) = 10

To FinD :

Sum of it's 10 terms

SolutioN :

We are given that 5th term of an AP is -3, common difference is -4 and we gave to find out sum of 10 terms.

First Term :

⇒an = a + (n - 1)d

⇒a5 = a + (5 - 1)-4

⇒-3 = a + 4(-4)

⇒-3 = a + (-16)

⇒-3 = a - 16

⇒a = -3 + 16

⇒a = 13

∴ First term of AP is 13.

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Sum of 10 terms :

⇒Sn = n/2 [2a + (n - 1)d]

⇒S10 = 10/2 [2(13) + (10 - 1)(-4)]

⇒S10 = 5 [26 + (9)(-4)]

⇒S10 = 5[26 - 36]

⇒S10 = 5(-10)

⇒S10 = -50

∴ Sum of 10 terms of AP is -50

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