Question

Solve $25+22+19+16+.....+x=115$

Answer 1

Hint: The given expression is an arithmetic progression. Find first term, common difference and last term. Using the formula of sum of n-terms. Substitute all the values and solve the quadratic equation formed to get the number of terms and solve for x.

Complete step-by-step answer:
The given expression, is in the form of an arithmetic progression,
$25+22+19+16+.....+x=115$
An arithmetic progression is a sequence of numbers such that the difference of any two successive members is constant. In the given expression (d) is the difference of 2^nd term to the first term
$d={{\text{2}}^{nd}}term-{{1}^{st}}term$
$=22-25=-3$
Similarly, $d={{3}^{rd}}term-{{2}^{nd}}term=19-22=-3$ .
Thus, we got the common difference as, $d=-3$
Now, the first term of the given sequence, $a=25$
We know that in AP, the formula to find ${{n}^{th}}$ term is ${{a}_{n}}=a+\left( n-1 \right)d$ …………………………….(1)
Let n be the number of terms in the given expression.
Put ${{a}_{n}}=x$ , as it is last term.
Hence, in equation (1) put ${{a}_{n}}=x$ , $a=25$ and $d=-3$ and simplify it.
$x=25+\left( n-1 \right)\left( -3 \right)$
$x=25-3n+3$
$x=28-3n$ ………………………………(2)
Now, we have been given the sum of n terms as 115.
Thus, we can say that ${{S}_{n}}=115$ .
We know the formula of AP, to find the sum of n terms as,
${{S}_{n}}=\dfrac{n}{2}\left( a+{{a}_{n}} \right)$
We know ${{S}_{n}}=115$ , $a=25$ and ${{a}_{n}}=x$
Thus, substitute three values and simplify it.
$115=\dfrac{n}{2}\left( 25+x \right)$
Put $x=28-3n$ in the above expression
$115=n\left( 25+28-3n \right)$
$230=n\left( 53-3n \right)$
$\Rightarrow 230=53n-3{{n}^{2}}$
$\therefore 3{{n}^{2}}-53n-230=0$
Now the above expression is similar to the quadratic formula, $a{{x}^{2}}+bx+c=0$. By comparing both the equation, we get $a=3,b=-53,c=230$
Thus, substitute these values in the quadratic formula,
$n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$n=\dfrac{-\left( -53 \right)\pm \sqrt{{{\left( -53 \right)}^{2}}-4\times 3\times 230}}{2\times 3}$
$=\dfrac{53\pm \sqrt{2809-2760}}{6}=\dfrac{53\pm \sqrt{49}}{6}=\dfrac{53\pm 7}{6}$
Thus, we have $n=\dfrac{53+7}{6}$ and $n=\dfrac{53-7}{6}=\dfrac{23}{3}$
                            $=\dfrac{60}{6}=10$
Hence, let us take $n=10$ , as the number of terms can’t be a fraction we neglect $n=\dfrac{23}{3}$.
Thus, we got the number of terms as $n=10$ .
Now put $n=10$ in equation (2) and get the value of x.
$x=28-3\times 10=28-30=-2$
Thus, we got $x=-2$ , which is the required solution.

Note: We can also find term x if we keep on subtracting 3 from the term,
$25+22+19+16+13+10+7+4+1-2=115$
Counting them we get a number of terms as 10 and the sum as 115.