Question

Tje value of the term is lying between the 10 th and 20 th term of an. Arithmetic progression is - 31 the value of the 20 th term is - 41 find the value of 10 th term​

Answer 1

Correct Question :- The value of the term is lying between the 10th and 20th term of an arithmetic progression is (- 31) . The value of the 20th term is (- 41) . find the value of 10th term ? ( All terms of AP are integers.)

Solution :-

As, we know, the nth term of an AP is given by :-

• T(n) = a + (n - 1)d
• a = first term of AP.
• d = common difference .

So,

→ T(20) = (-41)

→ a + (20 - 1)d = (-41)

→ a + 19d = (-41) ----------- Eqn.(1) .

Now, we have given that, the value of the term lying between the 10th term and 20th term is (-31).

Or,

→ from 11th to 19th term of given AP , one term is (-31).

Check :-

→ T(11) = a + (11 - 1)d = a + 10d .

if,

→ a + 10d = (-31).

Than, subtracting it from Eqn.(1) we get,

a + 19d = (-41)

a + 10d = (-31)

---------------------

9d = (-10)

---------------------

Here, we can see that, value of d is not an integer .

So, 11th term is Not Possible.

Similarly, checking next term now,

→ T(12) = a + (12 - 1)d = a + 11d = (-31)

subtracting it from Eqn.(1) we get,

a + 19d = (-41)

a + 11d = (-31)

---------------------

8d = (-10)

---------------------

again, value of d is not an integer .

So, Now, we gets our pattern :-

• we have to change value of term which is multiple of d , and if it divides the term (-10) completely, we gets values of d as an integer.
• 5d = (-10) => d = (-2) an integer.
• So, 19d - 5d = 14d .
• Than, a + 14d = a + (n - 1)d . => n = 15.

Therefore,

→ T(15) = (-31).

→ a + (n - 1)d = (-31)

→ a + 14d = (-31). ------------- Eqn.(2).

Subtracting Eqn.(2), from Eqn.(1) , we gets,

→ d = (-2) . { solved above. }

Putting value of d in Eqn.(1) ,

→ a + 19*(-2) = (-41)

→ a - 38 = (-41)

→ a = (-41) + 38

→ a = (-3).

Hence,

→ T(10) = a + (10 - 1)d

→ T(10) = (-3) + 9*(-2)

→ T(10) = (-3) - 18

→ T(10) = (-21) (Ans.)

∴ 10th term of given AP is (-21).