Question

nth term of AP is

$T_n$= (a-3)$n^3$+(b-4)$n^2$+(a+b)n-4

Then find:

1$^\text{st}$ common difference

2$^\text{nd}$ $a^2$+b²

3$^\text{rd}$ sum of first 10 terms

Answer 1

1st common difference = 7

2nd $a^2+b^2$ = 25

3rd sum of first 10 terms = 345

Answer 2

For the given sequence with $n$-th term $T_n = (a-3)n^3 + (b-4)n^2 + (a+b)n - 4$ to be an Arithmetic Progression (AP), the $n$-th term must be a linear function of $n$. This means the coefficients of $n^3$ and $n^2$ must be zero.

So, we must have: $a - 3 = 0 \implies a = 3$ $b - 4 = 0 \implies b = 4$

Substituting these values of $a$ and $b$ into the expression for $T_n$: $T_n = (3-3)n^3 + (4-4)n^2 + (3+4)n - 4$ $T_n = 0 \cdot n^3 + 0 \cdot n^2 + 7n - 4$ $T_n = 7n - 4$

This is the $n$-th term of an AP. The general form of the $n$-th term of an AP is $T_n = A + (n-1)D$, which can be written as $T_n = Dn + (A-D)$, where $A$ is the first term and $D$ is the common difference.

Comparing $T_n = 7n - 4$ with $T_n = Dn + (A-D)$: The coefficient of $n$ is the common difference $D$. So, the common difference is $D = 7$. The constant term is $A-D = -4$. Since $D=7$, we have $A - 7 = -4$, which gives $A = 3$. The first term is $T_1 = 7(1) - 4 = 3$, which matches $A=3$.

Now we can find the required values:

1. The first common difference: The common difference is $D = 7$.

2. $a^2 + b^2$: We found $a=3$ and $b=4$. $a^2 + b^2 = 3^2 + 4^2 = 9 + 16 = 25$.

3. Sum of the first 10 terms: The sum of the first $n$ terms of an AP is given by the formula $S_n = \frac{n}{2}(2A + (n-1)D)$. Here, $n=10$, the first term $A=3$, and the common difference $D=7$. $S_{10} = \frac{10}{2}(2 \cdot 3 + (10-1) \cdot 7)$ $S_{10} = 5(6 + 9 \cdot 7)$ $S_{10} = 5(6 + 63)$ $S_{10} = 5(69)$ $S_{10} = 345$.

Alternatively, using $S_n = \frac{n}{2}(A + T_n)$: First term $A = T_1 = 3$. 10th term $T_{10} = 7(10) - 4 = 70 - 4 = 66$. $S_{10} = \frac{10}{2}(T_1 + T_{10}) = 5(3 + 66) = 5(69) = 345$.