Question

Two APs have the same common difference. If the first terms of these APs be 3 and 8 respectively, find the difference between the sums of their first 50 terms.

Answer 1

Hey there !!

Given :-

→ a₁ = 8.

→ a'₁ = 3.

→ d₁ = d'₁ .

To find :-

→ S₅₀ - S'₅₀ .

Solution :-

Let a₁ and a'₁ be the two APs.

Then, a₁ = 8 and a'₁ = 3.

Let d be the common difference of the two APs.

Then,

→ S₅₀ - S'₅₀ .

[ Sn = n/2 ( 2a + ( n - 1 )d . ]

= [ 50/2 ( 2 × 8 + ( 50 - 1 )d ] - [ 50/2 ( 2 × 3 + ( 50 - 1 )d ] .

= [ 25 ( 16 + 49d ) ] - [ 25 ( 6  + 49d ].

= [ 400 + 1225d ] - [ 150 + 1225d ].

= 400 + 1225d - 150 - 1225d .

= 400 - 150 .

= 250.

Hence, 250 is the difference between the sum of their first 50 terms.

THANKS

#BeBrainly.

Answer 2

HEY MATE HERE IS YOUR ANSWER

$\Huge {\boxed {GIVEN :}}$

✔️ a1 = 8
✔️ a'1 = 3
✔️ d 1 = d' 1

We have to Find

${\Large{S 50 - S ' 50}}$

$\huge \bold {\boxed {Move \: To \: Solution \: :-}}$

Let a1 and a' 1 be the AP

➡️ a1 = 8 and a'1 = 3

And D is the common in two AP

Therefore :-
~~~~~~~~~~

➡️ S 50 - S '50

➡️ [ Sn = n/ 2 ( 2a + ( n - 1)d]

➡️ [50 /2 (2 × 8 + ( 50 - 1)d] - [50/2 ( 2× 3 + (50 - 1)d]

✔️ [25(16 + 49d)] - [ 25 ( 6 + 49d)]

✔️ 400 + 1225d - 150 - 1225d

➡️ 400 - 150

➡️ 250

Hence, $\underline {\boxed {250}}$ is the difference between the sum of their 50 terms.