Question

In an AP, s10=25,s11=26,then a11 is? ​

Answer 1

Answer:

Given:

• S10 = 25 and S11 = 26

To find:

• a11

Solution:

$\boxed{\sf{a_{n}=S_{n}-S_{n-1}}}$

$\sf{\therefore{a_{11}=S_{11}-S_{10}}}$

$\sf{From \ given}$

$\sf{\therefore{a_{11}=26-25}}$

$\sf{\therefore{a_{11}=1}} \\ \\ \sf{\therefore{The \ 11^{th} \ term \ of \ the \ A.P. \ is \ 1.}}$

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Extra information:

If a, b, and c are in A.P. we can also say that ak, bk and ck are in A.P. also we can say a/k, b/k and c/k are in A.P.

Answer 2

Solution

Given

⇒ S₁₀ = 25

⇒S₁₁ = 26

To find

⇒A₁₁

Formula

⇒Sₙ = n/2[2a + ( n - 1 )d ]

⇒Aₙ = a + ( n -1 )d

Now take

⇒S₁₀ = 25

⇒ 25 = 10/2[2a + ( 10 - 1 )d]

⇒25=5[2a + 9d]

⇒2a + 9d = 5                ....(i)eq

Now take

⇒S₁₁ = 26

⇒26 = 11/2[2a + (11 -1)d]

⇒52 = 11 [2a + 10d]

⇒52 = 22a + 110d         ...(ii)eq

Using substitution method

now take i eq

⇒2a + 9d = 5      

⇒2a = 5 - 9d

⇒ a = ( 5 - 9d )/2            ......(iii)

Now put the value of a on ii eq

⇒52 = 22a + 110    

∵ a = ( 5 - 9d )/2

⇒22(5-9d)/2+110d = 52

⇒11(5 -9d) + 110 d =52

⇒55 - 99d +110d = 52

⇒11d = 52 - 55

⇒d= -3/11

Put the value of d on iii eq

⇒ a = ( 5 - 9d )/2

⇒ a = ( 5 + 27/11)/2

⇒ a = ( 82/11 )/2

⇒a = 82/22=41/11

We have to find

⇒A₁₁

⇒Aₙ = a + ( n -1 )d

a=41/11 , d= -3/11

⇒A₁₁ = 41/11 + ( 11 - 1 ).-3/11

⇒A₁₁ = 41/11 - 30/11

⇒A₁₁ = 11/11 = 1