Test Index

CBSE Class 10 Standard Math 2026 All Sets Solved Paper

© examsnet.com
Question : 17 of 20
Marks: +1, -0
In an A.P., a = −3 and S17=357S_{17} = 357. The value of a17a_{17} is
Solution:  
Here, n = 17, a = -3, S17=357.S_{17} = 357.
Sn=  n2(a+l)S_n = \;\frac{n}{2} (a + l)
357=  172(−3+a17)357 = \;\frac{17}{2} (-3 + a_{17})
  357×217=−3+a17\;\frac{357 \times 2}{17} = -3 + a_{17}
21×2=−3+a1721 \times 2 = -3 + a_{17}
42=−3+a1742 = -3 + a_{17}
a17=42+3=45a_{17} = 42 + 3 = 45
The value of a17a_{17} is 45.
© examsnet.com
Go to Question:
Prev Question
Next Question