Question

The common difference of the whose Sn=3n square +7n is what

Answer 1

Answer:

Common difference = 6

Step-by-step explanation:

Hope you understood it

Answer 2

Question:

Find the common difference of the AP whose sum of first n terms is given as, S(n) = 3n² + 7n .

Answer:

Common difference, d = 6

Note:

• A sequence in which, the difference between the consecutive terms are same is called AP (Arithmetic Progression).

• Any AP is given as ; a , (a + d) , (a + 2d) , .....

• The nth term of an AP is given by ;

T(n) = a + (n - 1)•d , where a is the first term and d is the common difference of the AP .

• The common difference of an AP is given by ;

d = T(n) - T(n-1) .

• The sum of first n terms of an AP is given by ;

S(n) = (n/2)•[2a + (n-1)•d] .

• The nth term of an AP is given by ;

T(n) = S(n) - S(n-1) .

Solution:

We have;

S(n) = 3n² + 7n

Thus,

S(n-1) = 3(n-1)² + 7(n-1)

Now,

=> T(n) = S(n) - S(n-1)

=> T(n) = [3n² + 7n] - [3(n-1)² + 7(n-1)]

=> T(n) = 3n² + 7n - 3(n-1)² - 7(n-1)

=> T(n) = 3n² - 3(n-1)² + 7n - 7(n-1)

=> T(n) = 3[n² - (n-1)²] + 7[n - (n-1)]

=> T(n) = 3[n² - (n² - 2n + 1)] + 7(n - n + 1)

=> T(n) = 3[n² - n² + 2n - 1] + 7•1

=> T(n) = 3(2n - 1) + 7

=> T(n) = 6n - 3 + 7

=> T(n) = 6n + 4

Also,

T(n-1) = 6(n-1) + 4

Now,

=> d = T(n) - T(n-1)

=> d = [6n + 4] - [6(n-1) + 4]

=> d = 6n + 4 - 6(n-1) - 4

=> d = 6[n - (n-1)] + 4 - 4

=> d = 6[n - n + 1] + 0

=> d = 6•1

=> d = 6

Hence,

The common difference of the AP is 6 .