Question

please solve the question

Answer 1

a+ (p+1-1)d = 2[a+(q+1-1)d
a+pd = 2a+2qd
a =( p-2q)d
Now
a+ (3p+1-1) d = pd- 2qd + 3pd = 4pd-2qd ( lhs)
2[a+ ( p+q+1-1)d] = 2[pd-2qd+ pd + qd]
= 4pd-2qd ( rhs)
LHS = RHS

Answer 2

let first term of ap is 'a' & common diff is 'd'
now
Tp+1 = a+pd
Tq+1 = a+qd
a+pd= 2(a+qd) {given}.....(1)
now,
T3p+1=a+3pd =a+ pd+ 2pd
=2(a+qd)+2pd {using eqn (1)
=2{a+(p+q)d}=2{a+(p+q+1-1)d}
=2Tp+q+1 PROVED