Prove that a1 ,a2 ,…,an form an AP where an =3+4n Also find the sum of the first 15 terms of AP.
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 12
a1=3+4(1)=7
a2=3+4(2)=11
a2=3+4(3)=15
∴d=a3−a1=11−7=4
Here a=7, d=4 and n=15
By usingSn=2n[2a+(n−1)d] we have,
S15=215[2×7+(15−1)4]
=215(14+56)
=215×70=525