The first term of an AP is 14. The sum of the first five
terms and the first ten terms are equal in magnitude but
opposite in sign. Find the third term of that AP.
Answers
Answer :
a(3) = 70/11
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) .
Solution :
• Given : First term , a = 14 ; S(5) = – S(10)
• To find : Third term , a(3) = ?
We have ,
=> S(5) = - S(10)
=> (5/2)•[2a + (5-1)d] = - (10/2)•[2a + (10-1)d]
=> (5/2)•[ 2a + 4d ] = - (10/2)•[ 2a + 9d ]
=> (5/2)•2•[ a + 2d ] = - (10/2)•[ 2a + 9d ]
=> (10/2)•[ a + 2d ] = - (10/2)•[ 2a + 9d ]
=> a + 2d = - (2a+ 9d)
=> a + 2d = - 2a - 9d
=> 2d + 9d = -2a - a
=> 11d = -3a
=> d = -3a/11
=> d = -3×14/11
=> d = -42/11
Now ,
The third term of the AP will be ;
=> a(3) = a + (3 - 1)d
=> a(3) = a + 2d
=> a(3) = 14 + 2×(-42/11)
=> a(3) = 14 - 84/11
=> a(3) = (154 - 84)/11
=> a(3) = 70/11
Hence ,
Third term , a(3) = 70/11 .
Step-by-step explanation:
answer is 70/11.pls refer the solution attached
