Question

If the sum of first k terms of an A.P. is 3k2 - k
and its common difference is 6. What is the first
term?
​

Answer 1

Question: If the sum of first k terms of an A.P. is 3k² - k

and its common difference is 6. What is the first

term?

GIVEN:

Sum of first k terms of an AP = Sk = 3k² - k

Common Difference = 6

TO FIND:

First of the AP

SOLUTION:

We know that,

Sum of terms in an AP = n/2 [2a + (n - 1)d ]

3k² - k = k/2 [2a + (k - 1)d ]

3k² - k = k/2 [2a + (k - 1)6]

2(3k² - k) = k [2a + 6k - 6]

6k² - 2k = k [2a + 6k - 6]

(6k² - 2k)/k = 2a + 6k - 6

6k²/k - 2k/k = 2a + 6k - 6

6k - 2 = 2a + 6k - 6

6k - 6k - 2 = 2a - 6

=> -2 = 2a - 6

=> -2 + 6 = 2a

=> 4 = 2a

=> a = 4/2

=> a = 2

Therefore, the first term is 2.

Answer 2

Answer:

2

Step-by-step explanation:

Given :

$\displaystyle \sf S_k=3k^2-k$

Also common difference d = 6

Let first term be 'a' :

Using sum formula of A.P. :

$\displaystyle \sf 3k^2-k=\frac{k}{2}[ \ 2a+(k-1)6 \ ]$

$\displaystyle \sf 3k-1=\frac{1}{2}[ \ 2(a+(k-1)3 \ ]$

$\displaystyle \sf 3k-1=[ \ (a+(k-1)3 \ ]$

$\displaystyle \sf 3k-1=a+3k-3$

$\displaystyle \sf a=2$

Alternate method :

Given :

$\displaystyle \sf S_k=3k^2-k$

We know for first term , k = 1 :

$\displaystyle \sf a=3-1$

$\displaystyle \sf a=2$

Hence we get required answer.