Saved Bookmarks
| 1. |
Please Solve the question with steps . |
|
Answer» Hey mate here's your ANSWER and please MARK me as brainleast if you're happy with my answer. :) Step-by-step explanation: 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=>a=1 ∴T3=a |
|