Question

The 10th term and the 8th term of an A.P. are 25 and 41 respectively, then find the 38th term of that A.P.​

Answer 1

Answer :

-199

Note :

★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.

★ If a1 , a2 , a3 , . . . , an are in AP , then

a2 - a1 = a3 - a2 = a4 - a3 = . . .

★ The common difference of an AP is given by ; d = a(n) - a(n-1) .

★ The nth term of an AP is given by ;

a(n) = a + (n - 1)d .

★ If a , b , c are in AP , then 2b = a + c .

★ The sum of nth terms of an AP is given by ; S(n) = (n/2)×[ 2a + (n - 1)d ] .

or S(n) = (n/2)×(a + l) , l is the last term .

★ The nth term of an AP can be also given by ; a(n) = S(n) - S(n-1) .

★ A linear polynomial in variable n always represents the nth term of an AP .

★ A quadratic polynomial in variable n always represents the sum of n terms of an AP .

★ If each terms of an AP is multiplied or divided by same quantity , then the resulting sequence is an AP .

★ If same quantity is added or subtracted in each term of an AP then the resulting sequence is an AP .

Solution :

• Given : a(10) = 25 ; a(8) = 41
• To find : a(38) = ?

We have ;

=> a(10) = 25

=> a + (10 - 1)d = 25

=> a + 9d = 25 --------(1)

Also ,

=> a(8) = 41

=> a + (8 - 1)d = 41

=> a + 7d = 41 --------(2)

Now ,

Subtracting eq-(2) from (1) , we get ;

=> (a + 9d) - (a + 7d) = 25 - 41

=> a + 9d - a - 7d = -16

=> 2d = -16

=> d = -16/2

=> d = -8 -------(3)

Now ,

Using the formula , a(n) = a + (n-1)d

We have ;

=> a(38) = a + (38 - 1)d

=> a(38) = a + 37d

=> a(38) = (a + 7d) + 30d

=> a(38) = 41 + 30×(-8) {using eq-2,3}

=> a(38) = 41 - 240

=> a(38) = -199

Hence ,

The 38th term of the AP is -199 .

Answer 2

⚝ Given:-    

• ➛ a(10) = 25
• ➛ a(8) = 41

⚝ To find:-      

• ➛ a(38) =

⚝ Solution:-      

~Find out the 1st term(a) and common difference(d).

➳ a(10) = 25  

➳ a + (10 - 1)d = 25  

➳ a + 9d = 25 _____equation (1)

And,

➳ a(8) = 41  

➳ a + (8 - 1)d = 41  

➳ a + 7d = 41 ______equation (2)

~ Now, On Subtracting equations (1) and (2),

➟ (a + 9d) - (a + 7d) = 25 - 41  

➟ a + 9d - a - 7d = -16  

➟ 2d = -16  

➟ d = -16/2  

➟ d = -8 _________equation (3)

         _______________

Now we will use the formula,

• a(n) = a + (n-1)d

➼ a(n) = a + (n-1)d

➼ a(38) = a + (38 - 1)d  

➼ a(38) = a + 37d  

➼ a(38) = (a + 7d) + 30d  

➼ a(38) = 41 + 30×(-8) _____(use eq-2 and eq-3)  

➼ a(38) = 41 - 240  

➼ a(38) = -199

. ° . Thus, The 38th term of the AP is -199.

                 ____________

⚝ Answer:-    

• The 38th term of the AP is -199.