It is given that an = 3 – 4n, prove that a1 , a2 , a3 …… form an A.P. Also find S20.
Sir please give me a detailed solution of this question as it is very important for examination.
ML Aggarwal ( avichal publication)
Arithmetic Progression
Chapter 9 Question no 13
an=3−4n
a1=3−4(1)=−1
a2=3−4(2)=−5
a3=3−4(3)=−9
a3−a2=a2−a1=−4
as it should shows constant common difference
∴ It is an AP with a=−1 & d=−4
S20=220[2(−1)+(20−1)(−4)]
S20=10[−2−76]
S20=−780.