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

Answer:

hii

your answer is here !

Step-by-step explanation:

Given, first term, a = 10

Last term, al = 361

And, common difference, d = 9

Now al =a + (n −1)d

⟹ 361 = 10 + (n − 1)9

⟹ 361 = 10 + 9n − 9

⟹ 361 = 9n + 1

⟹ 9n = 360

⟹ n = 40

Therefore, total number of terms in AP = 40

Now, sum of total number of terms of an AP is given as:

Sn = n/2 [2a + (n − 1)d]

⟹ S40 = 40/2 [2 × 10 + (40 − 1)9]

= 20[20 + 39 x 9]

=20[20 + 351]

=20 × 371 = 7420

Thus, sum of all 40 terms of AP = 7420

follow me !

Answer 2

ANSWER:

Number of terms = 40

Sum of n terms = 7420.

GIVEN:

First term (a) = 10

Last term (l) = 361

Common difference (d) = 9

TO FIND:

Number of terms (n)

Sum of n terms in the AP.

SOLUTION:

Formula

$\implies \: last \: term(l) = a + (n - 1)d$

By using this formula we can calculate the number of terms.

$\implies \: 361 = 10 + (n - 1) \times 9 \\ \implies \: 361 - 10 = (n - 1) \times 9 \\ \implies \: 351 = (n - 1) \times 9 \\ \implies \: \frac{351}{9} = (n - 1) \\ \implies \: 39 = n - 1 \\ \implies \: n = 39 + 1 \\ \implies \: n = 40$

So number of terms(n) = 40

Now we have to find sum of 40(n) terms

formula

$sum \: (n \: terms) = \frac{n}{2} (2a + (n - 1)d$

Putting all the values we get;

$\implies \: sum = \frac{40}{2} (2 \times 10 + (40 - 1) \times 9 )\\ \implies \: sum = \: 20(20 + 351) \\ \implies \: sum = 20(371) \\ \implies \: sum = \: 7420$

Sum of 40(n) terms = 7420

NOTE:

some important formulas

$\implies \: sum \: (n \: terms) = \frac{n}{2} (2a + (n - 1)d \\ \implies \: nth \: term \: (beginning) = a + (n - 1)d \\ \implies \: nth \: term \: from \: last = l - (n - 1)d$

where. n = number of terms. a = first term

l = last term d= common difference