Question

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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.

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Answer 1

Answer:

→ 250 .

Step-by-step explanation:

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 .

[ $S_n$ = 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.

Answer 2

• Let two AP's be $a_{1}$ and $a_{1}'$

Where.. d is the common difference.

$a_{1}$ = 3 and $a_{1}'$ = 8 _______ [ GIVEN ]

» Two APs have the same common difference. If the first terms of these APs be 3 and 8.

We have to find the difference between the sums of their first 50 terms.

A.T.Q.

→ $S_{50}$ - $S_{50}'$

We know that..

$S_{n}$ = $\dfrac{n}{2}$ [2a + (n - 1)d]

So,

→ $S_{50}$ - $S_{50}'$

→ $\dfrac{50}{2}$ [2a + (50 - 1)d] - $\dfrac{50}{2}$ [2a + (50 - 1)d]

Put the known values

→ $\dfrac{50}{2}$ [2(8) + 49d] - $\dfrac{50}{2}$ [2(3) + 49d]

→ 25(16 + 49d) - 25(6 + 49d)

→ 400 + 1225d - 150 - 1225d

→ 250

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250 is the difference between the sums of their first 50 terms of AP's 3 and 8 respectively

________ [ ANSWER ]

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