Question

The 8th term of an AP is half its second term and the 11th term exceeds one
third of its 4th term by 1. Find its 15th term and the common difference of the
AP​

Answer 1

Step-by-step explanation:

ANSWER

Let a be the first term and d be the common difference of given AP.

Now, According to the question,

a8=1/2(a2)

and a11=1/3(a4)+1

⇒2×(a+7d)=a+d⇒a+13d=0 ...(1)

and 3×(a+10d)=(a+3d)+3⇒2a+27d=3 ...(2)

On applying (2)−2×(1), we get

d=3

Putting d=3 in (1), we get

a+13×3=0⇒a=−39

Now, a15=a+14d=−39+14×3=3

∴a15 =3

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Answer 2

Let a and d be the first term and the common difference of the A.P. respectively.

As per given,

a₈=1/2a₂

⇒a₂=2a₈

⇒(a+d)=2(a+7d) ( Using:aₙ=a+(n-1)d )

⇒a+d=2a+14d

⇒a+13d=0 ...(1)

It is also given that,

a₁₁=1/3a₄ + 1

⇒a₁₁=(a₄+3)/3

⇒3a₁₁=a₄+3

⇒3(a+10d)=a+3d+3 ( Using:aₙ=a+(n-1)d )

⇒3a+30d=a+3d+3

⇒2a+27d=3 ...(2)

Multiplying eq.(1) by 2, we get

2a+26d=0 ...(3)

Subtracting eq.(3) from eq.(2), we get

d=3

Putting d=3 in eq.(1), we get

a+13(3)=0

⇒a= -39

Now,

a₁₅=a+14d

⇒a₁₅= -39+14(3)

⇒a₁₅= 3