The perpendicular distance of point A (6, 9) from the x–axis and y-axis is
Answers
The number of trees in consecutive rows increase by 1. So, it this is an Arithmetic progression, where d = 1, a = trees in first row = 1 and n = number of rows = 25. We have to find out number of trees in 25 rows? Using well known formula, i.e, formula of sum of nth term of Arithmetic progression ::
\Large\underline{\boxed{\bf{\red{S_{n} = \dfrac{n}{2}\Big[2a + \big(n - 1\big)d\Big]}}}}
S
n
=
2
n
[2a+(n−1)d]
Where, Sn denotes sum of nth terms, n denotes number of terms, a denotes first term and d denotes common difference.
Let's solve it!!
\:
\underline{\sf{\bigstar\:Putting\:all\:known\:values\::-}}
★Puttingallknownvalues:−
\begin{gathered}\\ \longrightarrow \:\sf S_{25} = \dfrac{25}{2}\Big[\big(2\big)\big(1\big) + \big(25 - 1\big)\big(1\big)\Big] \end{gathered}
⟶S
25
=
2
25
[(2)(1)+(25−1)(1)]
\begin{gathered}\\ \longrightarrow \:\sf S_{25} = \dfrac{25}{2}\Big[\big(2\:\times\:1\big) + \big(24\:\times\:1\big)\Big] \end{gathered}
⟶S
25
=
2
25
[(2×1)+(24×1)]
\begin{gathered}\\ \longrightarrow \:\sf S_{25} = \dfrac{25}{2}\Big[2 + 24\Big] \end{gathered}
⟶S
25
=
2
25
[2+24]
\begin{gathered}\\ \longrightarrow \:\sf S_{25} = \dfrac{25}{\cancel{2}}\:\times\:\cancel{26}\end{gathered}
⟶S
25
=
2
25
×
26
\begin{gathered}\\ \longrightarrow \:\sf S_{25} = 25\:\times\:13\end{gathered}
⟶S
25
=25×13
\begin{gathered}\\ \longrightarrow \:\boxed{\bf {\purple{S_{25} = 325}}}\:\orange{\bigstar}\end{gathered}
⟶
S
25
=325
★