Question

If the sum of first 7 terms of an A.P. is 10 and that of next 7 terms is 17, find its 28th
term.​

Answer 1

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Answer 2

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Given:

If the sum of first 7 terms of an A.P. is 10 & that of next 7 terms is 17.

To find:

28th term.

Explanation:

We know that formula of the sum of A.P: Sn=$\frac{n}{2}$ [2a+(n-1)d]

We have,

• S7= 10
• S14 - S7= 17

$\bold{\underline{\underline{\huge{CASE.1}}}}}$

→ S7= $\frac{7}{2}$ [2a +(7-1)d]=10

→ S7= $\frac{7}{2}$ [2a+6d]= 10

→ S7= $\frac{7}{\cancel{2}}  [\cancel{2}a+\cancel{6}d]=10$

→ S7= 7[a+3d]=10

→ 7a +21d= 10......................(1)

$\bold{\underline{\underline{\huge{CASE.2}}}}}$

S14 - S7 = 17

→ $\frac{14}{2}$ [2a +(14-1)d] - $\frac{7}{2}$ [2a+(7-1)d]= 17

→ $\frac{14}{2}$ [2a+13d] - $\frac{7}{2}$ [2a +6d]=17

→ $\frac{28a+182d-14a -42d}{2} =17$

→ 28a -14a +182d-42d= 34

→ 14a + 140d =34

→ 2(7a +70d)=34

→ 7a +70d= 17........................(2)

Now,

Subtracting equation (1) from equation (2), we get;

⇒ \cancel{7a} +70d \cancel{-7a} -21d = 17-10

⇒ 70d - 21d = 7

⇒ 49d= 7

⇒ d= $\cancel{\frac{7}{49} }$

⇒ d= 1/7

Putting the value of d in equation (1),we get;

⇒ 7a +\cancel{21}× $\frac{1}{\cancel{7}}$ =10

⇒ 7a +3 = 10

⇒ 7a = 10 -3

⇒ 7a = 7

⇒ a= $\cancel{\frac{7}{7} }$

⇒ a= 1

We have,

• Common difference,d= 1/7

• First term,a= 1

T28 = 1+(28-1)× 1/7

T28 = 1 +27 × 1/7

T28 = $1+\frac{27}{7}$

T28= $\frac{7+27}{7}$

T28= $\frac{34}{7}$

Thus,

The 28th term of this A.P. is 34/7.