If the 9th term of an A.P is 0, prove that its 29th term is twice its 19th term

Mathematics

2 answers:

Ghella [55]3 years ago

6 0

Let the first term, common difference and number of terms of an AP are a, d and n respectively.

Given that, 9th term of an AP, T9 = 0 [∵ nth term of an AP, Tn = a + (n-1)d]

⇒ a + (9-1)d = 0

⇒ a + 8d = 0 ⇒ a = -8d ...(i)

Now, its 19th term , T19 = a + (19-1)d

= - 8d + 18d [from Eq.(i)]

= 10d ...(ii)

and its 29th term, T29 = a+(29-1)d

= -8d + 28d [from Eq.(i)]

= 20d = 2 × T19

Hence, its 29th term is twice its 19th term

Scilla [17]3 years ago

5 0

Answer:

Proved below.

Step-by-step explanation:

a9 = a1 + 8d = 0 where a1 = first term and d = common difference.

we need to prove that

a1 + 28d = 2(a1 + 18d

simplifying:-

a1 + 36d - 28d = 0

a1 + 8d = 0 which is what we are given.

Therefore the proposition is true.

You might be interested in

How to solve -d/6+12=-7

Delicious77 [7]

IF you are solving for d:

isolate the D, do the opposite of PEMDAS.

-d/6 + 12 = -7

(subtract 12 from both sides)

-d/6 + 12 (-12) = -7 (-12)

-d/6 = -19

(multiply 6 to both sides)

-d/6(6) = -19(6)

-d = -19(6)

-d = -144

-d/-1 = -144/-1

d = 144

hope this helps

6 0

3 years ago

Write the point-slope form of the equation of the horizontal line that passes through the point (2, 1). Pease show how!

mote1985 [20]

Check the picture below.

using a second for it, since it's a horizontal line, then say hmmm we'll use the y-intercept, or (0, 1), so the equation of a line that passes through (2,1) and (0,1)

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 2}}\quad ,&{{ 1}})\quad 
% (c,d)
&({{ 0}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-1}{0-2}\implies \cfrac{0}{-2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=0(x-2)$