Animated Solution for Mathematics - Sequence and Series: Let $a_1,a_2,.........,a_{30}$ be an A. P., $S={\sum }_{i=1}^{30}{a}_i$ and $T={\sum }_{i=1}^{15}{a}_{(2i-1)}$. If $a_5=27$ and $S-2T=75$, then $a_{10}$ is equal to :

Imagine you are standing at the start of a long, perfectly spaced line of numbers. This is an Arithmetic Progression (AP), a sequence where every step you take is of the exact same size, $d$.

We are given an AP with 30 terms, $a_1,a_2,\dots ,a_{30}$. We have two sums: $S$, the sum of all 30 terms, and $T$, the sum of only the odd-indexed terms. Our goal is to find $a_{10}$.

The standard formula for the sum of the first $n$ terms of an AP is:

For our total sum $S$, we have $n=30$. Substituting this, we get:

Expanding this, we obtain our first pillar:

Now, let us look at $T=a_1+a_3+a_5+\cdots +a_{29}$. Notice that $T$ is itself an arithmetic progression.

The first term is $a_1$, the common difference is $2d$, and there are exactly 15 terms. Applying the sum formula:

Simplifying the expression inside the brackets:

To facilitate our calculation, we compute $2T$:

We are given the relationship $S-2T=75$. Substituting our derived expressions:

The $a_1$ terms cancel out entirely, leaving:

Dividing by 15, we find the common difference:

Now that we have $d$, we use the given condition $a_5=27$. Using the general term formula $a_n=a_1+(n-1)d$:

Substituting $d=5$:

Finally, we calculate the target term $a_{10}$:

The final answer is 52.