Question

Difficulty: MediumArithmetic and Geometric Progressions (AP and GP)

The 4th4^{\text{th}} term of an arithmetic progression (A.P.) is 1515 and the 9th9^{\text{th}} term is 3535. Calculate the sum of the first 1010 terms of the progression.

Answer: 210

Answer

The sum of the first 10 terms of the progression is 210.
Using the nthn^{\text{th}} term formula Tn=a+(n1)dT_n = a + (n-1)d, the equations a+3d=15a + 3d = 15 and a+8d=35a + 8d = 35 yield common difference d=4d = 4 and first term a=3a = 3. Substituting these into the sum formula Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n-1)d] for n=10n = 10 yields S10=5[2(3)+9(4)]=210S_{10} = 5[2(3) + 9(4)] = 210.

Step-by-Step Solution

1
Set up equations for the given terms
a+3d=15a + 3d = 15 and a+8d=35a + 8d = 35
The nthn^{\text{th}} term of an A.P. is given by Tn=a+(n1)dT_n = a + (n-1)d.
2
Solve for the common difference dd
d=4d = 4
Subtracting (a+3d=15)(a + 3d = 15) from (a+8d=35)(a + 8d = 35) gives 5d=205d = 20, so d=4d = 4.
3
Solve for the first term aa
a=3a = 3
Substituting d=4d = 4 into a+3(4)=15a + 3(4) = 15 yields a=1512=3a = 15 - 12 = 3.
4
Calculate the sum of the first 10 terms
S10=210S_{10} = 210
Applying S10=102[2(3)+9(4)]=5(6+36)=210S_{10} = \frac{10}{2}[2(3) + 9(4)] = 5(6 + 36) = 210.

Key Concept

Finding terms and sums of an Arithmetic Progression using simultaneous linear equations
Estimated Time:1m 30s
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