Question

There are ten numbers in a certain A.P. The sum of first three terms is 321. The sum of last
three numbers is 405. Find the sum of all the ten numbers.
a) 1165
b) 1210
c) 1221
d) 1252
plz answer with explaination!!!!!!!!!!!urgent!!!!!!!!!!!!!​

Answer 1

$\:\:\:\: \: \: \: \: \large\mathfrak{\underline{\huge\mathcal{\bf{\boxed{\boxed{\huge\mathcal{~~QUESTION~~}}}}}}}$

There are ten numbers in a certain A.P. The sum of first three terms is 321. The sum of lastthree numbers is 405. Find the sum of all the ten numbers.

a) 1165

b) 1210

c) 1221

d) 1252

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \large\mathfrak{\underline{\huge\mathcal{\bf{\boxed{\boxed{\huge\mathcal{~~AnsweR~~}}}}}}}$

_____________________________________________

$<font color=red><marquee behavior=alternate>Solution</marquee></font>$

let the first term is=a

common difference=d

now sum of first 3 terms..is..

$= > \frac{3(2a + (3 - 1)d)}{2} = 321 \\ = > a + d = 107$

now sum of last 3 terms..is..

$= > a(8) + a(9) + a(10) = 405 \\ = > a + (8 - 1)d + a + (9 - 1)d +a + (10 - 1)d = 405 \\ = > 3a + 24d = 405 \\ = > a + 8d = 135$

a +8d = 135

a + d = 107

━━━━━━━━━

7d=28

=>d=4

and .... a=107-4=103

therefore....

$s(10) = \frac{10(2 \times 103 + (10 - 1)3)}{2} \\ = \frac{2330}{2} \\ = 1165$

option (a)1165

$<font color=green><marquee direction="Right">hope this help you$

Answer 2

The sum of all the ten numbers of the AP is 1210.

b is the correct option.

Step-by-step explanation:

Let the given AP is

a, a+d, a+2d,...........+a+7d, a+8d, a+9d

Now, it has been given that, sum of first three terms is 321. Thus, we have

a+a+d+a+2d = 321

3a+3d=321

3(a+d)=321

a+d = 107............(i)

Now, second condition is "sum of last  three numbers is 405". Therefore, we have

a+7d+a+8d+a+9d=405

3a+24d=405

3(a+8d)=405

a + 8d = 135.......(ii)

Subtract equation (i) from (ii)

$7d=28\\\\d=\frac{28}{7}\\\\d=4$

Substituting this value in (i)

a + 4 = 107

a = 107 - 4

a = 103

Therefore, sum of 10 terms of this AP is given by

$S=\frac{n}{2}(2a+(n-1)d)\\\\=\frac{10}{2}(2\cdot103+(10-1)4)\\\\=5(206+36)\\\\=1210$

B is the correct option.

#Learn More:

In an A.P., the sum of first ten terms is −150 and the sum of its next ten terms is −550. Find the A.P.

https://brainly.in/question/10761082

The sum of first six terms of an A.P. is 69 and the  sum of last three terms of the same A.P. is 273.  If the first term of the A.P. is 4, find the number of  terms in that ​A.P​

https://brainly.in/question/10692998