(a_5=11), (a_9=17)
[ a+4d=11,\quad a+8d=17 ] [ 4d=6\implies d=\dfrac{3}{2},\quad a=11-4\cdot\dfrac{3}{2}=11-6=5 ]
[ S_{20}=\dfrac{20}{2}\bigl[2\cdot5+19\cdot\dfrac{3}{2}\bigr]=10\bigl(10+\dfrac{57}{2}\bigr)=10\cdot\dfrac{77}{2}=385 ]
Answer: (385)
11TH•Mathematics•Sequences And Series•Question 8
Overview
The 5th and 9th terms of an A.P. are 11 and 17 respectively. Find the sum of 20 terms.
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