Question

In an Arithmetic progression,

$\sf t_{15}=30,t_{20}=50$
Find,
$\sf S_{17}$​

Answer 1

Question : -

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

Find S₁₇ ?

Answer

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

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;

S_(nth) = (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 ✓

Answer 2

QuestioN :-

$\sf \: in \: an \: arithmetic \: progression \\ \\ \sf \: t _{15} = 30 \:, \: \: t _{20} = 50 \\ \\ \sf \: find \\ \\ \implies \sf s_{17} = \: ?$

Given :-

• $\sf \: t _{15} = 30 \: \: , \: \: t _{20} = 50 \:$

AnsweR :-

In the given terms ,

$\sf \: t_{15} \: represent \: the \: 15th \: element \: of \: AP \:$

$\sf \: t_{20} \: represent \: the \: 20th \: element \: of \: AP \:$

We know that :-

• $\sf \: t_{15} \: can \: be\: represented \: as \: = a + 14d$
• $\sf \: t_{20} \: can \: be\: represented \: as \: = a + 19d$

Now,

$\star \sf \: a + 14d = 30 - - - - (i) \\ \\ \star\sf \: a + 19d = 50 - - - - (ii)$

Subtracting the equation (i) from equation (ii) →

$\implies \sf \: (a + 19d) - (a + 14d) = 50 - 30 \\ \\ \: \implies \sf \: a + 19d - a - 14d = 20 \\ \\ \implies \: \sf \: 5d = 20 \\ \\ \implies \sf \: d = \frac{20}{5} \\ \\ \implies \boxed{\sf \: d = 4} \\ \\ \sf \star \: hence \: common \: difference \: (d) = 4$

Now , putting the value of d into Equation (i) →

$\implies \sf \: a + 14(4) = 30 \\ \\ \implies \sf \: a + 56 = 30 \\ \\ \implies \: \sf \: a = 30 - 56 \\ \\ \implies \boxed{ \sf \: a = - 26} \\ \\ \sf \star \: hence \: \: first \: term(a) = - 26$

Now, Find out the sum of first 17 terms →

We know that :-

• $\sf \: s_{n} = \frac{n}{2} \{2a + (n - 1)d \}$

But , n = 17

$\implies \sf \: s_{17} = \frac{17}{2} \{2 \times( - 26) + (17 - 1)4 \} \\ \\ \implies \sf \: s_{17} = \frac{17}{2} \{ - 52 + 16 \times 4\ \\ \\ \implies \sf \: s_{17} = \frac{17}{2} \{ - 52 + 64\} \\ \\ \implies \sf \: s_{17} = \frac{17}{ \cancel{2}} \times \cancel{12} \\ \\ \implies \sf \: s_{17} = 17 \times 6 \\ \\ \implies \boxed{\sf \: s_{17} = 108} \:$