Question

The first and the last terms of an AP are 10 and 361 respectively. If its common difference is 9 then find the number of terms and their total sum?​

Answer 1

$\Huge{\underline{\underline{\mathfrak{\red{Question :}}}}}$

The first and the last terms of an AP are 10 and 361 respectively. If its common difference is 9 then find the number of terms and their total sum?

$\Huge{\underline{\underline{\mathfrak{\orange{Solution :}}}}}$

Last term of A.P :- 361

First term of A.P :- 10

Difference :- 9

A.P :- 10, 19, 28, 37 .................. 361

We know that,

$\LARGE \implies {\boxed{\boxed{\blue{\sf{{a_{n} = a + (n - 1)d}}}}}}$

(Putting Values)

361 = 10 + (n - 1)9

⟹ 361 - 10 = (n - 1)9

⟹ 351 = (n - 1)9

⟹ 351/9 = n - 1

⟹ 39 = n - 1

⟹ 39 + 1 = n

⟹ 40 = n

$\Huge{\boxed{\green{\sf{n = 40}}}}$

$\rule{200}{2}$

Now,

$\LARGE \implies {\boxed{\boxed{\pink{\sf{S_{n} =\frac{n}{2} a + L}}}}}$

Putying Values

Sn = 40 (10 + 361)/ 2

⟹ Sn = 40 (371) / 2

⟹ Sn = 14840 / 2

⟹ Sn = 7420

$\Huge{\boxed{\purple{\sf{S_{n} = 7420}}}}$

Answer 2

Solution :

Given :

• First term of an AP a = 10
• Last term of an an AP = 361
• Common difference d = 9

$\text{Let } a_n = 361$

By using nth term of an AP formula

$\tt a_n = a + (n - 1)d$

Substituting the values

⇒ 361 = 10 + (n - 1)9

⇒ 361 - 10 = (n - 1)9

⇒ 351 = (n - 1)9

⇒ 351/9 = n - 1

⇒ 39 = n - 1

⇒ 39 + 1 = n

⇒ 40 = n

⇒ n = 40

Hence, there are 40 terms in AP.

Now, using Sum of terms of an AP formula

$\tt S_n = \dfrac{n}{2} (a + l)$

Here

• First term a = 10
• Last term l = 361
• Number of terms n = 40
• Sum of terms Sₙ = ?

Substituting the values

$\implies S_n = \dfrac{40}{2} (10+ 361)$

$\implies S_n = 20 \times 371$

$\implies S_n = 7420$

Hence, their total sum is 7420.