Question

The first and last terms of an AP are 7 and 49 respectively. If sum of all terms is 420. Find its common difference ?​

Answer 1

Answer:

The first and the last terms of an A.P are 7 and 49, respectively. If the sum of all its terms is 420, find its common difference. ... Hence the common difference of the A.P = 3. Note: [a] This question can also be solved using the formula Sn=n2(2a+(n−1)d).

Answer 2

Given : First term of an A.P ( or Airtgmetic Progression ) is : 7 & Let Term of an A.P ( or Airtgmetic Progression ) is : 49 & Sum of all terms of an A.P is 420.

To Find : Find the common difference of an A.P ?

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Solution : Let's Consider the number of terms be n & the common difference be d.

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$\underline{\frak{As~ we ~know ~that~:}}$

• $\boxed{\sf\pink{S^{n}~=~\dfrac{n}{2}\bigg(a~+~l\bigg)}}$

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Where,

• S_n is the sum of all terms of A.P
• n is the number of terms in A.P
• l is the last term of an A.P
• a is the first term of an A.P

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$\pmb{\sf{\underline{Substituting~ the~ Given ~Values~:}}}$

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$\qquad:\implies {\sf{ S^n =\: \dfrac{n}{2} \bigg( a + l \bigg)}}$

$\qquad:\implies{ \sf{ 420 =\: \dfrac{n}{2} \bigg( 7 + 49 \bigg)}}$

$\qquad:\implies{ \sf {420 =\: \dfrac{n}{2} \bigg( 56 \bigg)}}$

$\qquad:\implies{ \sf{ 420 \times 2 =\: n \bigg( 56 \bigg)}}$

$\qquad:\implies {\sf {840 =\: n \bigg( 56 \bigg)}}$

$\qquad:\implies {\sf{ 840 =\: 56n}}$

$\qquad:\implies {\sf {\cancel{\dfrac{840}{56}} =\: n}}$

$\qquad:\implies {\sf {15 =\: n}}$

$\qquad:\implies{\underline{\boxed{\frak{\purple{n~=~15}}}}}$★

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Therefore,

• The last term is 15.

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$\qquad :\implies{ \sf{ a + ( n - 1 ) d \:=\: 49}}$

$\qquad :\implies{ \sf {7 + (15 - 1 ) d \:=\: 49}}$

$\qquad :\implies{ \sf {7 + (14 ) d \:=\: 49}}$

$\qquad :\implies{ \sf {7 + 14 d \:=\: 49}}$

$\qquad:\implies {\sf {14 d \:=\: 49 - 7}}$

$\qquad :\implies{ \sf {14 d \:=\: 42}}$

$\qquad :\implies{\sf{ d \:=\: \cancel{\dfrac{42}{14}}}}$

$\qquad:\implies{\underline{\boxed{\frak{\pink{d~=~3}}}}}$★

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Hence,

$\underline{\sf{The~common~difference~or~\bf{\underline{d}}~\sf{is}~\bf{\underline{3}}}}$