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