Question

ifthe first and last term of an Ap are 7 and 49 respectively. If the sum of all its terms is 402,find its common difference

Answer 1

Correct question: If the first and last terms of A.P are 7 and 49 respectively. If the sum of all its terms is 420, find its common difference.

Answer:

$\large{\underline{\boxed{\sf D = 3}}}$

Step-by-step explanation:

Given that;

• a₁ = 7
• last term = 49
• Sum of all terms = 420

We know that;

Sₙ = $\sf \dfrac{n}{2}$ [a + l]

⇒ 420 = $\sf \dfrac{n}{2}$ [7 + 49]

⇒ 840 = n * 56

⇒ n = $\sf \dfrac{840}{56}$

⇒ n = 15

Also, we know ;

aₙ = a + (n - 1)d

⇒ 49 = 7 + (15 - 1)d

⇒ 14d = 49 - 7

⇒ 14d = 42

⇒ d = $\sf \dfrac{42}{14}$

⇒ d = 3

Hence, the common difference is 3.

Answer 2

HeRe Is Your Ans ⤵

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Correct Question :-

➡If The First and last term of an AP are 7 and 49 respectively. If the sum of all its terms is 420 , Find its common difference.

Ans :-

➡Common Difference = 3

Given :-

➡First Term = 7

➡Last Term = 49

➡Sum = 420

To Find :-

➡Commom Difference

Solution :-

$= > S _{n} = \frac{n}{2} (a + l) \\ \\ = > 420 = \frac{n}{2} ( 7 + 49) \\ \\ = > 420 = \frac{n}{2} \times 56 \\ \\ = > n = \frac{840}{56} \\ \\ = > n = \fbox{15}$

$= > A _{n} = a + (n - 1)d \\ \\ = > 49 = 7 + (15 - 1)d \\ \\ = > 49 = 7 + 14d \\ \\ = > d = \frac{42}{14} \\ \\ = > d = 3$

Hence ，Common Difference Of Given AP Is $\fbox{3}$

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