Question

find the sum of 1st 25termsofAP 7,11,15,19.....by using formula​

Answer 1

Answer:-

Given:

a (first term) = 7

d (Common difference) = t(2) - t(1)

d = 11 - 7

d = 4

n = 25

We know that,

S(n) = n/2 * [ 2a + (n - 1)d ]

→ S(25) = 25/2 * [ 2(7) + (25 - 1)(4) ]

→ S(25) = 25/2 * [ 14 + 24(4) ]

→ S(25) = 25/2 * (14 + 96)

→ S(25) = (25*110)/2

→ S(25) = 1375

Hence, the sum of first 25 terms of the given AP is 1375.

Additional Information:-

• nth term of an AP = a + (n - 1)d

• Arithmetic mean of a , b , c => (a + c)/2 = b.

• Sum of first n natural numbers = n (n + 1)/2.

• Sum of squares of first n natural numbers = n(n + 1)(2n + 1)/6

• Sum of cubes of first n natural numbers = n²(n + 1)²/4

• Sum of cubes of first n natural numbers = (Sum of first n natural numbers)².

Answer 2

☯ GiveN :

• A.P : 7, 11, 15, 19
• First term (a) = 7
• Common Difference (d) = 4
• Number of terms (n) = 25

$\rule{200}{1}$

☯ To FinD :

We have to find the sum of first 25 terms of the A.P.

$\rule{200}{1}$

☯ SolutioN :

We know the formula to calculate the sum of terms.

$\Large{\implies{\boxed{\boxed{\sf{S_n = \frac{n}{2} \bigg(2a + (n - 1)d \bigg)}}}}}$

★ Putting Values ★

$\sf{\dashrightarrow S_{25} = \frac{25}{2} \bigg(2(7) + (25 - 1)4 \bigg)} \\ \\ \sf{\dashrightarrow S_{25} = \frac{25}{2} \bigg(14 + (24)4 \bigg)} \\ \\ \sf{\dashrightarrow S_{25} = \frac{25}{2} (14 + 96)} \\ \\ \sf{\dashrightarrow S_{25} = \frac{25}{\cancel{2}} \times \cancel{110}} \\ \\ \sf{\dashrightarrow S_{25} = 25 \times 55} \\ \\ \sf{\dashrightarrow S_{25} = 1375} \\ \\ \Large{\implies{\boxed{\boxed{\sf{S_{25} = 1375}}}}} \\ \\ \sf{\therefore \: Sum \: of \: first \: 25 \: terms \: of \: the \: A.P \: is \: 1375}$