25, 22, 19,... There are 22 terms find the Sum of the AP. 2) An A.p has 15 terms and a Common difference of…

Question

25, 22, 19,... There are 22 terms find the Sum of the AP. 2) An A.p has 15 terms and a Common difference of −3 find its last term its Sum is 120 3) 2, 4, 8, 16,... There are 16 terms find the Sum of the AP 4) 8, 12, 18,... There are 6 terms find the Sum of the G.P

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Here are the solutions to the problems:

1) For the Arithmetic Progression (AP) 25, 22, 19, ... with 22 terms:

Identify the first term ( $a$ ), common difference ( $d$ ), and number of terms ( $n$ ).
The first term is $a = 25$ .
The common difference is $d = 22 - 25 = -3$ .
The number of terms is $n = 22$ .
Use the formula for the sum of an AP, $S_n = \frac{n}{2}[2a + (n-1)d]$ .
$S_{22} = \frac{22}{2}[2(25) + (22-1)(-3)]$ $S_{22} = 11[50 + (21)(-3)]$ $S_{22} = 11[50 - 63]$ $S_{22} = 11[-13]$ $S_{22} = -143$
The sum of the AP is $\boxed{-143}$ .

2) For an AP with 15 terms, common difference of -3, and a sum of 120:
Identify the given values and use the sum formula to find the first term ( $a$ ).
Given $n = 15$ , $d = -3$ , $S_{15} = 120$ .
The formula for the sum of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$ .
$120 = \frac{15}{2}[2a + (15-1)(-3)]$ $120 = \frac{15}{2}[2a + (14)(-3)]$ $120 = \frac{15}{2}[2a - 42]$
Multiply both sides by $\frac{2}{15}$ :
$\frac{120 \times 2}{15} = 2a - 42$ $16 = 2a - 42$ $16 + 42 = 2a$ $58 = 2a$ $a = \frac{58}{2}$ $a = 29$
The first term is $a = 29$ .
Use the formula for the $n$ -th term of an AP, $a_n = a + (n-1)d$ , to find the last term ( $a_{15}$ ).
$a_{15} = 29 + (15-1)(-3)$ $a_{15} = 29 + (14)(-3)$ $a_{15} = 29 - 42$ $a_{15} = -13$
The last term is $\boxed{-13}$ .

3) For the sequence 2, 4, 8, 16, ... with 16 terms:
Identify the type of progression and its parameters.
The sequence 2, 4, 8, 16, ... is a Geometric Progression (GP), not an AP, because each term is multiplied by a constant ratio to get the next term.
The first term is $a = 2$ .
The common ratio is $r = \frac{4}{2} = 2$ .
The number of terms is $n = 16$ .
Use the formula for the sum of a GP, $S_n = \frac{a(r^n - 1)}{r - 1}$ (since $r > 1$ ).
$S_{16} = \frac{2(2^{16} - 1)}{2 - 1}$ $S_{16} = \frac{2(2^{16} - 1)}{1}$ $S_{16} = 2(65536 - 1)$ $S_{16} = 2(65535)$ $S_{16} = 131070$
The sum of the GP is $\boxed{131070}$ .

4) For the Geometric Progression (GP) 8, 12, 18, ... with 6 terms:
Identify the first term ( $a$ ), common ratio ( $r$ ), and number of terms ( $n$ ).
The first term is $a = 8$ .
The common ratio is $r = \frac{12}{8} = \frac{3}{2} = 1.5$ .
The number of terms is $n = 6$ .
Use the formula for the sum of a GP, $S_n = \frac{a(r^n - 1)}{r - 1}$ (since $r > 1$ ).
$S_6 = \frac{8\left(\left(\frac{3}{2}\right)^6 - 1\right)}{\frac{3}{2} - 1}$ $S_6 = \frac{8\left(\frac{3^6}{2^6} - 1\right)}{\frac{1}{2}}$ $S_6 = 16\left(\frac{729}{64} - 1\right)$ $S_6 = 16\left(\frac{729 - 64}{64}\right)$ $S_6 = 16\left(\frac{665}{64}\right)$ $S_6 = \frac{16 \times 665}{64}$ $S_6 = \frac{665}{4}$ $S_6 = 166.25$
The sum of the GP is $\boxed{166.25}$ .

25, 22, 19,... There are 22 terms find the Sum of the AP. 2) An A.p has 15 terms and a Common difference of -3 find its last term its Sum is 120 3) 2, 4, 8, 16,... There are 16 terms find the Sum of the AP 4) 8, 12, 18,... There are 6 terms find the Sum of the G.P

Step-by-step solution

Identify the first term ( $a$ ), common difference ( $d$ ), and number of terms ( $n$ ).

Use the formula for the sum of an AP, $S_n = \frac{n}{2}[2a + (n-1)d]$ .

Identify the given values and use the sum formula to find the first term ( $a$ ).

Use the formula for the $n$ -th term of an AP, $a_n = a + (n-1)d$ , to find the last term ( $a_{15}$ ).

Identify the type of progression and its parameters.

Use the formula for the sum of a GP, $S_n = \frac{a(r^n - 1)}{r - 1}$ (since $r > 1$ ).

Identify the first term ( $a$ ), common ratio ( $r$ ), and number of terms ( $n$ ).

Use the formula for the sum of a GP, $S_n = \frac{a(r^n - 1)}{r - 1}$ (since $r > 1$ ).

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More Mathematics Questions

25, 22, 19,... There are 22 terms find the Sum of the AP. 2) An A.p has 15 terms and a Common difference of -3 find its last term its Sum is 120 3) 2, 4, 8, 16,... There are 16 terms find the Sum of the AP 4) 8, 12, 18,... There are 6 terms find the Sum of the G.P

Step-by-step solution

Identify the first term (aaa), common difference (ddd), and number of terms (nnn).

Use the formula for the sum of an AP, Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d]Sn​=2n​[2a+(n−1)d].

Identify the given values and use the sum formula to find the first term (aaa).

Use the formula for the nnn-th term of an AP, an=a+(n−1)da_n = a + (n-1)dan​=a+(n−1)d, to find the last term (a15a_{15}a15​).

Identify the type of progression and its parameters.

Use the formula for the sum of a GP, Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1}Sn​=r−1a(rn−1)​ (since r>1r > 1r>1).

Identify the first term (aaa), common ratio (rrr), and number of terms (nnn).

Use the formula for the sum of a GP, Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1}Sn​=r−1a(rn−1)​ (since r>1r > 1r>1).

Still stuck on this one?

More Mathematics Questions

Identify the first term ( $a$ ), common difference ( $d$ ), and number of terms ( $n$ ).

Use the formula for the sum of an AP, $S_n = \frac{n}{2}[2a + (n-1)d]$ .

Identify the given values and use the sum formula to find the first term ( $a$ ).

Use the formula for the $n$ -th term of an AP, $a_n = a + (n-1)d$ , to find the last term ( $a_{15}$ ).

Use the formula for the sum of a GP, $S_n = \frac{a(r^n - 1)}{r - 1}$ (since $r > 1$ ).

Identify the first term ( $a$ ), common ratio ( $r$ ), and number of terms ( $n$ ).

Use the formula for the sum of a GP, $S_n = \frac{a(r^n - 1)}{r - 1}$ (since $r > 1$ ).