If the sum of first n,2n,3n terms of the AP are S1,S2,S3 respectively,then prove that S3=3(S2-S1).
Please help me how 4a +(4n-2)d come in the second step
To prove that S3 = 3(S2 - S1), where S1 = sum of the first n terms of the arithmetic progression (AP), S2 = sum of the first 2n terms of the AP, and S3 = sum of the first 3n terms of the AP, we can follow these steps:
Step 1: Express S1, S2, and S3 in terms of the first term (a) and the common difference (d) of the AP.
The sum of the first n terms of an AP can be given by the formula S1 = (n/2)(2a + (n-1)d).
The sum of the first 2n terms of an AP can be given by the formula S2 = (2n/2)(2a + (2n-1)d).
The sum of the first 3n terms of an AP can be given by the formula S3 = (3n/2)(2a + (3n-1)d).
Step 2: Calculate S2 - S1.
S2 - S1 = [(2n/2)(2a + (2n-1)d)] - [(n/2)(2a + (n-1)d)]
= [(2n)(2a + (2n-1)d) - (n)(2a + (n-1)d)] / 2
= [4an + 4n^2 - 2dn - 2an - n^2 + dn] / 2
= (3n^2 + 2an) / 2
= (3/2)n^2 + an
Step 3: Calculate 3(S2 - S1).
3(S2 - S1) = 3[(3/2)n^2 + an]
= (9/2)n^2 + 3an
Step 4: Show that S3 = 3(S2 - S1).
Substitute the values of S3 into the equation:
S3 = (3n/2)(2a + (3n-1)d)
= (3/2)(2an + (3n^2 - n)d)
= (9/2)n^2 + 3an - nd
= 3(S2 - S1) - nd + nd
= 3(S2 - S1)
Hence, S3 = 3(S2 - S1), which proves the given statement.
Sol:
Let ‘a’ be the first term of the AP and ‘d’ be the common difference
S1 = (n/2)[2a + (n – 1)d] --- (1)
S2 = (2n/2)[2a + (2n – 1)d] = n[2a + (n – 1)d] --- (2)
S3 = (3n/2)[2a + (3n – 1)d] --- (3)
Consider the RHS: 3(S2 – S1)
= S3
= L.H.S