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In an Arithmetic progression,

t15​=30,t20​=50
Find,
S17​=??


User Chalda
by
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1 Answer

10 votes

Question : -

In an Arithmetic progression,
\sf{t_(15) = 30 \; , t_(20) = 50 }

Find S₁₇ ?

Given :-

  • t₁₅ = 30; t₂₀ = 50

Required to find : -

  • S₁₇=??

Solution : -

Given that;

  • t₁₅ = 30; t₂₀ = 50

t₁₅ represents the 15th term of the AP

t₂₀ represents the 20th term of the AP

Now,

We know that;

t₁₅ can be represented as , a + 14d

t₂₀ can be represented as , a + 19d

Now, This implies

  • a + 14d = 30 ••••••(1)

  • a + 19d = 50 •••••••(2)

Subtract eq 1 from eq 2

a + 19d = 50

a + 14d = 30

(-)(-) (-)

-------------------

0 + 5d = 20

  • 5d=20-0
  • 5d = 20
  • d = (20)/(5)
  • d = 4

So,

Common difference (d) = 4

Substituting the value of d in eq-1

  • a + 14d = 30
  • a + 14(4) = 30
  • a + 56 = 30
  • a = 30 - 56
  • a = -26

So,

First term (a) = -26

Now,

Let's find the sum of first 17 terms !

We know that;


\tt{S_(n)= (n)/(2)[2a+(n-1)d]}

Now,

Here the no. of term (n) = 17

We have,

  • S₁₇ = (17)/(2) [2(-26)+(17-1)4]
  • S₁₇ = (17)/(2) [-52 + (16)4]
  • S₁₇ = (17)/(2) [-52 + 64]
  • S₁₇ = (17)/(2) [12]
  • S₁₇ = (17 x 12)/(2)
  • S₁₇ = 17 x 6
  • S₁₇ = 108

Therefore,

S₁₇ = 108

User Vreal
by
5.1k points