Question

ANSWER WITH EXPLANATION TO BE THE BRAINLIEST
Find the common difference of an A.P. where first term is 100 and the sum of whose first 6 terms is 5 times.The sum of the next 6 terms

A) 10 B) –5 C) 6 D) –10

Answer 1

Answer:

( D )

Step-by-step explanation:

From the properties of AP :

 nth term is given by a + ( n - 1 )d and sum of first n terms is given by (n/2){ 2a + ( n - 1 )d }, where a is the first term and d is the common difference between the terms.

  Therefore, let the common difference between the terms be d

Sum of first 6 terms should be

⇒ ( 6 / 2 ){ 2( 100 ) + ( 6 - 1 )d }

⇒ 3( 200 + 5d )

     As d is the common difference :  

 a₇ = a + 6d = 100 + 6d

 a₁₂ = a + 11d = 100 + 11d

Let us consider a new AP( in that AP ) with first term as 100 + 6d and last term as 100 + 11d.

Thus, the sum of next 6 terms we will use (n/2)( a + l ) where a is the first term and l is the last term.

 So, sum of next 6 terms is :

⇒ ( 6 / 2 )( 100 + 6d + 100 + 11d )

⇒ 3( 200 + 17d )

Given,

  Sum of whose first 6 terms is 5 times the sum of the next 6 terms

⇒ 3( 200 + 5d ) = 5[ 3( 200 + 17d ) ]

⇒ 200 + 5d = 5( 200 + 17d )

⇒ 40 + d = 200 + 17d

⇒ 40 - 200 = 17d - d

⇒ - 160 = 16d

⇒ - 160 / 16 = d

⇒ - 10 = d

    Hence, common difference between the terms is - 10.

Answer 2

Answer:

$\bigstar\:\boxed{\sf S_n=\dfrac{n}{2}\Bigg\lgroup2a+(n-1)d\Bigg\rgroup}$

Let the First Term be a and Common Difference be d of the AP.

$:\implies\sf S_6=\dfrac{6}{2}\Bigg\lgroup2a+(6-1)d\Bigg\rgroup\\\\\\:\implies\sf S_6=3\big\lgroup2a + 5d\big\rgroup\\\\\\:\implies\sf S_6 = 6a + 15d\\\\\\\\:\implies\sf S_{12}=\dfrac{12}{2}\Bigg\lgroup2a+(12-1)d\Bigg\rgroup\\\\\\:\implies\sf S_{12}=6\big\lgroup2a + 11d\big\rgroup\\\\\\:\implies\sf S_{12} = 12a + 66d$

$\rule{150}{1}$

$\underline{\bigstar\:\textbf{According to the Question :}}$

$:\implies\textsf{Sum of first 6 terms = 5 \big\lgroup Sum of the next 6 terms \big\rgroup }\\\\\\:\implies\sf S_6=5\big\lgroup S_{12}-S_6\big\rgroup\\\\\\:\implies\sf 6a + 15d = 5\big\lgroup 12a + 66d- 6a -15d\big\rgroup\\\\\\:\implies\sf 6a + 15d = 5\big\lgroup6a + 51d\big\rgroup\\\\\\:\implies\sf 6a + 15d = 30a + 255d\\\\\\:\implies\sf 6a - 30a = 255d - 15d\\\\\\:\implies\sf - \:24a = 240d\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Putting the value of a = 100.}}\\\\:\implies\sf - \:24(100) = 240d\\\\\\:\implies\sf -\:2400=240d\\\\\\:\implies\sf \dfrac{-\:2400}{24}=d\\\\\\:\implies\underline{\boxed{\sf d=-\:10}}$

⠀

$\therefore\:\underline{\textsf{Common Difference of AP will be D) \textbf{- 10}}}.$