Question

1 -
The sum of the first nine terms of an arithmetic progression is 117. The sum of the next four terms is 91. Find the first term and the common difference of the progression.

Answer 1

Answer:

The sum of first 30 terms of Arithmetic Progression is 3105

Step-by-step explanation:

Given data

Sum of first 6 terms of arithmetic progression 

Sum of first 12 terms of arithmetic progression 

To find - the sum of first 30 terms of arithmetic progression 

The formula to find the sum of first n terms of arithmetic progression is

Where n is the number of terms

a is the first term of the progression

d is the common difference

For the sum of first 6 terms,

39 = 2 a +5 d -----> (1)

For the sum of first 12 terms ,

81 = ( 2 a + 11 d ) -----> (2)

Subtract (2) from (1)

we get, 6 d = 42

              d = 7

Substitute the value of 'd' in equation (2)

2 a + 11 (7) = 81

2 a + 77 = 81

2 a = 81 - 77

2 a = 4

a = 2

Then for the Sum of first 30 terms of arithmetic progression is

Therefore the sum of first 30 terms of the arithmetic progression is 3105 when the sum of first 6 terms is 117 and sum of first 12 terms is 486.

Step-by-step explanation:

I hope it will be help you

Answer 2

Answer:

Step-by-step explanation:

Well it's quite simple if you go over it for a while;

Initially, we may derive the following equation

9/2 (2a + 8d) = 117

Secondly, we are given that the next 4 terms are equivalent to 91

Therefore, we can construct a second equation

We'll start it with 9d as the next 4 terms are the 10th, 11th, 12th, and the 13th terms.

(a + 9d) + (a + 10d) + (a + 11d) + (a + 12d) = 91

Simplify it to 4a + 42d = 91

Now we have both of our equations, we can simplify them;

4a + 42d = 91

9/2 (2a + 8d) = 117 or 2a + 8d = 26

Simplify for either a or b

we get D as 3/2 or 1.5

and A as 7

Hope you understood! :D