Question

The A.P. in which fourth term is - 15 and 9th term -30. find the sum of first 15 numbers. (write answer in decimals)​

Answer 1

Solution:-

Given :-

$\rm\implies \: T_4 = - 15$

$\rm \implies \: T_9 = - 30$

To find

$\rm \implies \: S_{15}$

Formula

$\to \rm \: T_n = a + (n - 1)d$

$\rm \to S_n = \dfrac{n}{2} \{2a + (n - 1)d \}$

Now

$\rm\implies \: T_4 = - 15$

$\to \rm \: - 15 = a + (4 - 1)d$

$\to \rm \: a + 3d = - 15 \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)$

$\rm \implies \: T_9 = - 30$

$\to \rm \: -30 = a + (9 - 1)d$

$\to \rm \: a + 8d = - 30 \: \: \: \: \: \: \: \: \: \: \: .......(ii)$

Now subtract (i) from (ii)

$\rm \: a + 3d - (a + 8d)= - 15 - ( - 30)$

$\rm \: a + 3d - a - 8d = 15$

$\rm \: - 5d = 15$

$\rm \: d = - 3$

put the value on (i) eq

$\to \rm \: a + 3 \times - 3 = - 15 \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)$

$\rm \: a - 9 = - 15$

$\rm \: a = - 15 + 9$

$\rm \: a = - 6$

Now we have to find

$\rm \implies \: S_{15}$

we have

$\rm \: a = - 6$

$\rm \: d = - 3$

$\rm \: n = 15$

Formula

$\rm \to S_n = \dfrac{n}{2} \{2a + (n - 1)d \}$

use the formula we get

$\rm \to S_{15} = \dfrac{15}{2} \{2 \times - 6 + (15 - 1) \times - 3 \}$

$\rm \to S_{15} = \dfrac{15}{2} \{ - 12 + (14) \times - 3 \}$

$\rm \to S_{15} = \dfrac{15}{2} \{ - 12 - 42 \}$

$\rm \to S_{15} = \dfrac{15}{2} \{ 54 \}$

$\rm \to S_{15} = \dfrac{15}{2} \times 54$

$\rm \to S_{15} = {15}{} \times 27$

$\rm \to S_{15} = 405$

Answer 2

Answer:

$Solution:-</p><p></p><p>Given :-</p><p></p><p>\rm\implies \: T_4 = - 15⟹T4=−15</p><p></p><p>\rm \implies \: T_9 = - 30⟹T9=−30</p><p></p><p>To find</p><p></p><p>\rm \implies \: S_{15}⟹S15</p><p></p><p>Formula</p><p></p><p>\to \rm \: T_n = a + (n - 1)d→Tn=a+(n−1)d</p><p></p><p>\rm \to S_n = \dfrac{n}{2} \{2a + (n - 1)d \}→Sn=2n{2a+(n−1)d}</p><p></p><p>Now</p><p></p><p>\rm\implies \: T_4 = - 15⟹T4=−15</p><p></p><p>\to \rm \: - 15 = a + (4 - 1)d→−15=a+(4−1)d</p><p></p><p>\to \rm \: a + 3d = - 15 \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)→a+3d=−15...(i)</p><p></p><p>\rm \implies \: T_9 = - 30⟹T9=−30</p><p></p><p>\to \rm \: -30 = a + (9 - 1)d→−30=a+(9−1)d</p><p></p><p>\to \rm \: a + 8d = - 30 \: \: \: \: \: \: \: \: \: \: \: .......(ii)→a+8d=−30.......(ii)</p><p></p><p>Now subtract (i) from (ii)</p><p></p><p>\rm \: a + 3d - (a + 8d)= - 15 - ( - 30)a+3d−(a+8d)=−15−(−30)</p><p></p><p>\rm \: a + 3d - a - 8d = 15a+3d−a−8d=15</p><p></p><p>\rm \: - 5d = 15−5d=15</p><p></p><p>\rm \: d = - 3d=−3</p><p></p><p>put the value on (i) eq</p><p></p><p>\to \rm \: a + 3 \times - 3 = - 15 \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)→a+3×−3=−15...(i)</p><p></p><p>\rm \: a - 9 = - 15a−9=−15</p><p></p><p>\rm \: a = - 15 + 9a=−15+9</p><p></p><p>\rm \: a = - 6a=−6</p><p></p><p>Now we have to find</p><p></p><p>\rm \implies \: S_{15}⟹S15</p><p></p><p>we have</p><p></p><p>\rm \: a = - 6a=−6</p><p></p><p>\rm \: d = - 3d=−3</p><p></p><p>\rm \: n = 15n=15</p><p></p><p>Formula</p><p></p><p>\rm \to S_n = \dfrac{n}{2} \{2a + (n - 1)d \}→Sn=2n{2a+(n−1)d}</p><p></p><p>use the formula we get</p><p></p><p>\rm \to S_{15} = \dfrac{15}{2} \{2 \times - 6 + (15 - 1) \times - 3 \}→S15=215{2×−6+(15−1)×−3}</p><p></p><p>\rm \to S_{15} = \dfrac{15}{2} \{ - 12 + (14) \times - 3 \}→S15=215{−12+(14)×−3}</p><p></p><p>\rm \to S_{15} = \dfrac{15}{2} \{ - 12 - 42 \}→S15=215{−12−42}</p><p></p><p>\rm \to S_{15} = \dfrac{15}{2} \{ 54 \}→S15=215{54}</p><p></p><p>\rm \to S_{15} = \dfrac{15}{2} \times 54→S15=215×54</p><p></p><p>\rm \to S_{15} = {15}{} \times 27→S15=15×27</p><p></p><p>\rm \to S_{15} = 405→S15=405</p><p></p><p>$