Question

find the sum of first 23 term of an ap whose n term is 5 - 2n​

Answer 1

Answer:

Construction- Extend the line segment DE and produce it to F such that, EF = DE.

In triangle ADE and CFE,

EC = AE —– (given)

∠CEF = ∠AED (vertically opposite angles)

∠DAE = ∠ECF (alternate angles)

By ASA congruence criterion,

△ CFE ≅ △ ADE

Therefore,

∠CFE = ∠ADE {by c.p.c.t.}

∠FCE= ∠DAE {by c.p.c.t.}

and CF = AD {by c.p.c.t.}

∠CFE and ∠ADE are the alternate interior angles.

Assume CF and AB as two lines which are intersected by the transversal DF.

In a similar way, ∠FCE and ∠DAE are the alternate interior angles.

Assume CF and AB are the two lines which are intersected by the transversal AC.

Therefore, CF ∥ AB

So, CF ∥ BD

and CF = BD {since BD = AD, it is proved that CF = AD}

Thus, BDFC forms a parallelogram.

By the properties of a parallelogram, we can write

BC ∥ DF

and BC = DF

BC ∥ DE

and DE = (1/2 * BC).

Hence, the midpoint theorem is proved.

Answer 2

Answer: Sum of its first 20 terms of AP is 320.

Step-by-step explanation:

Given: an=5−2n

Find some of the terms of AP:

Put n=1, we get a1=5-(2*1)=5-2=3 first term

Put n=2, we get a2=5-(2*2)=5-4=1 second term

Common difference=d=a2−a 1

​ =1-3=−2

Sum of first n terms:

Sn =n/2[2a+(n−1)d]

Sum of first 20 terms:

S20=20/2[2(3)+(20−1)(−2)]

=10[−6+19*−2]

=10[−6+38]

=10[32]

=320

Sum of its first 20 terms of AP is 320