Question

if the first term of an A.P is -5 and the last term is 25 the total number of term is 10 then find the sum of all terms​

Answer 1

$\huge\sf\pink{Answer}$

☞ Your Answer is 100

━━━━━━━━━━━━━

$\huge\sf\blue{Given}$

✭ First term of an AP is -5

✭ Last term of the AP is 25 and there are a total of 10 terms

━━━━━━━━━━━━━

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

◈ Sum of all the 10 terms of the AP?

━━━━━━━━━━━━━

$\huge\sf\purple{Steps}$

The nth term of an AP is given by,

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

◕ a = First term

◕ n = nth term of the AP

◕ d = Common Difference

We shall use this formula to find the common differce of the AP

➝ $\sf a_n = a+(n-1)d$

➝ $\sf a_{10} = -5+(10-1)d$

➝ $\sf 25 = -5+9d$

➝ $\sf 25+5 = 9d$

➝ $\sf 30 = 9d$

➝ $\sf \dfrac{30}{9} = d$

➝ $\sf \red{d = \dfrac{10}{3}}$

Now the sum of the nth term is given by,

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

Substituting the given values,

➳ $\sf S_{10} = \dfrac{10}{2}\bigg\lgroup 2(-5) +(10-1)(\dfrac{10}{3})\bigg\rgroup$

➳ $\sf S_{10} = 5\bigg\lgroup -10+ \cancel{9}\times \dfrac{10}{\cancel{3}}\bigg\rgroup$

➳ $\sf S_{10} = 5\bigg\lgroup -10+3(10)\bigg\rgroup$

➳ $\sf S_{10} = 5(20)$

➳ $\sf \orange{S_{10} = 100}$

━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

given= a=-5, l= 25,n=10

now,

sn=n/2(a+l)

=10/2(-5+25)

=5(20)=100

sn =100

Step-by-step explanation:

hope this will help u