Question

solve this question plz plz plz.....
xd​

Answer 1

$\bf\large\green{\underline{ given : }}$

• Sum of 1st 7 terms of AP is 49.
• Sum of next 10 terms of AP is 240.

$\\ \bf\large\green{\underline{to \: find : }}$

• Sum of 1st 'n' terms.

$\\ \bf\large\green{\underline{ formula \: used : }}$

$\boxed{\bf { a(n) = a + (n - 1)d }} \\ \\ \boxed{\bf{s(n) = \frac{n}{2} (1st \: term \: + \: last \:term }}$

Here ,

• a(n) ➡ 'n' th term.
• a ➡ 1st term.
• n ➡ total number of terms.
• d ➡ common difference.
• s(n) ➡ sum of 'n' terms.

$\bf\large\green{\underline{solution : \: }}$

∵ Sum of first 7 terms is 49.

$\sf\green{\underline{ }}s(7) = \frac{n}{2} (a + (a + 6d)) \\ \\ \sf\green{\underline{ }}49 = \frac{7}{2} (2a + 6d) \\ \\ \sf\green{\underline{ }}14 = 2a + 6d \\ \\ \sf\red{ 7 = a + 3d } \: \: \: \: \: \: \: \: - - - (1)$

∵ Sum of next 10 terms is 240.

∴ Sum of first 17 terms will be (240+49) = 249.

$\sf\green{\underline{ }}s(17) = \frac{17}{2} (a + (a + 16d)) \\ \\ \sf\green{\underline{ }}289 = \frac{17}{2} (2a + 16d) \\ \\ \sf\green{\underline{ }}34 = 2a + 16d \\ \\ \sf\red{17 = a + 8d } \: \: \: \: \: \: \: - - - (2)$

Substracting equation (1) and (2) ,

• a + 8d - a - 3d = 17 - 7
• 5d = 10

d = 2

Putting this value in equation (1),

7 = a + 3d

7 = a + 3(2)

7 = a + 6

a = 1

Sum of 'n' terms,

$\sf\green{\underline{ }}s(n) = \frac{n}{2} (a + (a + (n - 1)d) \\ \\ = > \sf\green{\underline{ }}s(n) = \frac{n}{2} (1 + (1 + (n - 1)2) \\ \\ = > \sf\green{\underline{ }}s(n) = \frac{n}{2} (2 + 2n - 2) \\ \\ = > \sf\green{\underline{ }} s(n) = \frac{n}{2} (2n) \\ \\ \bf\pink{\underline{ s(n) = {n}^{2} }}$

∴ Sum of 'n' numbers is n².

Answer 2

$\bf\large{\underline{ given : }}$

Sum of 1st 7 terms of AP is 49.

Sum of next 10 terms of AP is 240.

$\\ \bf\large{\underline{to \: find : }}$

Sum of 1st 'n' terms.

$\\ \bf\large{\underline{ formula \: used : }}$

$\boxed{\bf { a(n) = a + (n - 1)d }} \\ \\ \boxed{\bf{s(n) = \frac{n}{2} (1st \: term \: + \: last \:term }}$

Here ,

a(n) ➡ 'n' th term.

a ➡ 1st term.

n ➡ total number of terms.

d ➡ common difference.

s(n) ➡ sum of 'n' terms.

$\bf\large{\underline{solution : \: }}$

∵ Sum of first 7 terms is 49.

$\sf{\underline{ }}s(7) = \frac{n}{2} (a + (a + 6d)) \\ \\ \sf\green{\underline{ }}49 = \frac{7}{2} (2a + 6d) \\ \\ \sf\green{\underline{ }}14 = 2a + 6d \\ \\ \sf\red{ 7 = a + 3d } \: \: \: \: \: \: \: \: - - - (1)$

∵ Sum of next 10 terms is 240.

∴ Sum of first 17 terms will be (240+49) = 249.

$\sf{\underline{ }}s(17) = \frac{17}{2} (a + (a + 16d)) \\ \\ \sf\green{\underline{ }}289 = \frac{17}{2} (2a + 16d) \\ \\ \sf\green{\underline{ }}34 = 2a + 16d \\ \\ \sf\red{17 = a + 8d } \: \: \: \: \: \: \: - - - (2)$

Substracting equation (1) and (2) ,

a + 8d - a - 3d = 17 - 7

5d = 10

d = 2

Putting this value in equation (1),

7 = a + 3d

7 = a + 3(2)

7 = a + 6

a = 1

Sum of 'n' terms,

$\sf{\underline{ }}s(n) = \frac{n}{2} (a + (a + (n - 1)d) \\ \\ = > \sf{\underline{ }}s(n) = \frac{n}{2} (1 + (1 + (n - 1)2) \\ \\ = > \sf{\underline{ }}s(n) = \frac{n}{2} (2 + 2n - 2) \\ \\ = > \sf{\underline{ }} s(n) = \frac{n}{2} (2n) \\ \\ \bf\red{\underline{ s(n) = {n}^{2} }}$

∴ Sum of 'n' numbers is n².