Question

question of 10 class's maths​

Answer 1

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) & 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, a

15

=a+14d=−39+14×3=3

∴a15=3

Step-by-step explanation:

please mark me as brainliest ❤️ ❤️❤️❤️❤️❤️ please

Answer 2

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

Now, According to the question,

a8=1/2(a2) & 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, a

15

=a+14d=−39+14×3=3

∴a15=3