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The sum of the first 6 terms of an arithmetic progression (AP) is 42. The ratio of its 11th term to its 33rd term is 1:3. Calculate the first term of the AP.

Answer :

The first term of the arithmetic progression is 2.

Let's denote the first term of the arithmetic progression (AP) as \( a \), and the common difference as ( d ).

1. Using the Sum of the First 6 Terms of the AP:

The sum of the first ( n ) terms of an arithmetic progression can be calculated using the formula:

[tex]\[ S_n = \frac{n}{2} [2a + (n - 1)d] \][/tex]

Given that the sum of the first 6 terms is 42, we can write:

[tex]\[ 42 = \frac{6}{2} [2a + (6 - 1)d] \][/tex]

[tex]\[ 42 = 3 [2a + 5d] \][/tex]

[tex]\[ 14 = 2a + 5d \][/tex]

[tex]\[ 2a + 5d = 14 \] ...(Equation 1)[/tex]

2. Using the Ratio of the 11th and 33rd Terms:

The nth term of an AP can be calculated using the formula:

[tex]\[ T_n = a + (n - 1)d \][/tex]

Given that the ratio of the 11th term to the 33rd term is 1:3, we can write:

[tex]\[ \frac{T_{11}}{T_{33}} = \frac{a + (11 - 1)d}{a + (33 - 1)d} = \frac{1}{3} \][/tex]

[tex]\[ 3(a + 10d) = a + 32d \][/tex]

[tex]\[ 3a + 30d = a + 32d \][/tex]

[tex]\[ 2a = 2d \][/tex]

[tex]\[ a = d \][/tex] ...(Equation 2)

Now, we have two equations:

1) [tex]\( 2a + 5d = 14 \)[/tex]

2) [tex]\( a = d \)[/tex]

Let's solve these equations simultaneously to find the value of \( a \):

From Equation 2, substitute \( a = d \) into Equation 1:

[tex]\[ 2(a) + 5(a) = 14 \][/tex]

[tex]\[ 7a = 14 \][/tex]

[tex]\[ a = \frac{14}{7} \][/tex]

[tex]\[ a = 2 \][/tex]

So, the first term of the arithmetic progression is 2.

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