Question

The 5th term of an AP is -3 and its common difference is -4. The sum of its first 10 terms is__________.

Answer 1

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

$\therefore$ First term of AP is 13.

_______________________________

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

$\therefore$Sum of 10 terms of AP is -50

Answer 2

$\huge\sf\pink{Answer}$

☞ $\sf S_{10} = -50$

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ The 5th term of anAP is -3

✭ Common Difference (d) = -4

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

☆ The sum of the first 10 terms of the AP?

$\rule{110}1$

$\huge\sf\purple{Steps}$

We know that,

➝ $\sf a_n = a + (n-1)d$

➝ $\sf a_5 = a + (5-1)(-4)$

➝ $\sf -3 = a + (-16)$

➝ $\sf -3 = a -16$

➝ $\sf -3+16 = a$

➝ $\sf\red{a = 13}$

So now the sum of the first 10 terms is,

➳ $\sf S_n = \dfrac{n}{2}\bigg\lgroup 2a+(n-1)d\bigg\rgroup$

➳ $\sf S_{10} = \dfrac{10}{2}\bigg\lgroup 2(13)+(10-1)(-4)\bigg\rgroup$

➳ $\sf S_{10} = 5 \bigg\lgroup 26+ (9)(-4)\bigg\rgroup$

➳ $\sf S_{10} = 5\bigg\lgroup 26-36\bigg\rgroup$

➳ $\sf\orange{ S_{10} =-50}$

$\rule{170}3$