Question

the sum of all terms of the ap having 10 terms is 99 and the except for the sixth term is 89. find the third term of the progression if the sum of the first term and the fifth term is equal to 10.

Answer 1

Answer:

Given:

S10=99+T1..........(i)S10=89+T6..........(ii)

where S10 is the sum of 10 terms of the A.P. and T1,T6 are the first and sixth term respectively.

Say a and d are the first term and common difference of the A.P. respectively.

∴S10=5{2a+9d};T1=a;T6=a+5d........(iii)∴5{2a+9d}=a+99........(iv)5{2a+9d}=a+89+5d........(v)

Subtracting (iv) and (v), we get,

10−5d=0=>d=2........(vi)

Also given that

T1+T5=10=>a+a+4d=10=>2a+4×2=10=>2a=2=>

Step-by-step explanation:

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