Question

7th term of ap is 128 9th term of ap 140 find
the 11th term​

Answer 1

EXPLANATION.

7th term of an A.P. = 128.

9th term of an A.P. = 140.

As we know that,

General term of an A.P.

⇒ Tₙ = a + (n - 1)d.

⇒ T₇ = a + (7 - 1)d = 128.

⇒ a + 6d = 128. ⇒ (1).

⇒ T₉ = a + (9 - 1)d = 140.

⇒ a + 8d = 140. ⇒ (2).

From equation (1) & (2), we get.

⇒ a + 6d = 128.

⇒ a + 8d = 140.

We get,

⇒ - 2d = - 12.

⇒ d = 6.

Put the value of d = 6 in equation (1), we get.

⇒ a + 6(6) = 128.

⇒ a + 36 = 128.

⇒ a = 128 - 36.

⇒ a = 92.

To find : T₁₁ term.

⇒ T₁₁ = a + (11 - 1)d.

⇒ T₁₁ = a + 10d.

⇒ T₁₁ = 92 + 10(6).

⇒T₁₁ = 92 + 60.

⇒ T₁₁ = 152.

                                                                                                                   

MORE INFORMATION.

Supposition of terms in A.P.

(1) = Three terms as : a - d, a, a + d.

(2) = Four terms as : a - 3d, a - d, a + d, a + 3d.

(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: \: answer \: \: is \: \: as \: \: follows:-}}}}}$

★GIVEN:-

• 7th term of an AP = 128

• 9th term of an AP = 140

★TO FIND:-

• 11th term

★SOLUTION:-

In 7th term =

➝ a + 6d = 128 ---------------- (1.)

In 9th term =

➝ a + 8d = 140 ---------------- (2.)

Subtract (2.) - (1.)

‎ ‎a + 8d = 140

- ‎a + 6d = 128

----------------------

⟹ 2d = 12

So,

$: \implies \sf{d = \frac{12}{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \boxed { \underline{ d \: = 6}}}$

Put the value of d in eq. (1.)

➝ a + 8d = 140

➝ a + 8 (6) = 140

➝ a = 140 - 48

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \boxed{ \underline{ \sf{a = 92}}}}$

Now, we can find 11th term

➝ 11th term = a + 10 d

➝ 11th term = 92 + 10 × 6

${ \boxed{ \underline{ \color{teal}{ \sf{ \bf{11th \: \: term = 152}}}}}}$

So, the 11th term is 152.

━━━━━━━━━━━━━━━━━━✿

★MORE TO KNOW:-

• $\sf{a_n = a_1 + (n-1)d}$

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

• $\sf{S_n = \frac{n}{2} (a + a_n)}$

• $\sf{A.P. = a, a+d, a + 2d, a + 3d,....}$

HERE:-

• an = The nth term in the sequence.

• a1 = the first term in the sequence.

• d = the common difference between terms.