Question

In an ap if the 12th term is13 and sum of its first four terms is 24 find the sum of its first 10 terms

Answer 1

Step-by-step explanation:

a12 = 13 => a+12d ------> 1

a=13-12d

s4 = 24 = 4/2(2a +3d) ------->2

substituting a=13-12d in 2eq, we get

2(2×13-12d + 3d)=24

=> 12 = 26 -24d +3d

=> 12 = 26 - 21d

=> d = -14/-21

=> d = 14/21

Answer 2

The sum of the first 10 terms is 0.

Step-by-step explanation:

Hi there,

There is a slight misprint in the question given. I have written the question correctly and solved it accordingly. Thanks.

Q. If 12th term of ap is -13 and sum of first 4 terms is 24 find sum of first 10 terms

Step 1:

Let the first term in an A.P. be "a" and the common difference be "d".

It is given that,

12th term of an A.P. is -13 i.e., a₁₂ = -13.

We have  the formula for the nth term of an A.P. as,

aₙ = a + (n-1)d

substituting the values in the formula,

a₁₂ = a + (12 - 1)d

⇒ - 13 = a + 11d ......... (i)

Step 2:

Also, sum of first four terms is 24 i.e., S₄ = 24

We have the formula for the sum of first 4 terms as,

Sₙ =  $\frac{n}{2}$ [2a + (n-1)d]

substituting the values in the formula,

S₄ = $\frac{4}{2}$[2a + (4 - 1)d]

⇒ 24 = 2 [2a + 3d]

⇒ 12 = 2a + 3d ....... (ii)

Now, multiplying eq. (i) by 2 and then subtracting eq. from it, we get

- 26 = 2a + 22d

12    = 2a + 3d

-         -      -

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

 - 38 = 19d

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

∴ d = - 38/19 = -2

substituting value of d in eq. (i)

-13 = a + 11×(-2)

⇒ -13 = a - 22

⇒ a = 22 - 13

⇒ a = 9

Step 3:

Thus,

The sum of its first 10 terms i.e.,

S₁₀ =  $\frac{10}{2}$ [2×(9) + (10 - 1)(-2)] = 5 [18 - 18)] = 5 × 0 = 0

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

Also View:

Find the sum of first 25 terms of the AP 3,9/2,6,15/2,....

https://brainly.in/question/6541644

The 20th term of the arithmetic progression 10,7,4,........ is option provided

a)-57 b)47 c)-47 d)67

https://brainly.in/question/8927728